Redes de Comunicación Conceptos Fundamentales y Arquitecturas Básicas 1ra Edicion Alberto Leon Garcia

SIGUENOS EN:
LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS GRATIS EN
DESCARGA DIRECTA
VISITANOS PARA DESARGALOS GRATIS.
Solutions to Chapter 3
1. Suppose the size of an uncompressed text file is 1 megabyte.
Solutions follow questions: [4 marks – 1 mark each for a & b, 2 marks c]
a. How long does it take to download the file over a 32 kilobit/second modem?
T32k = 8 (1024) (1024) / 32000 = 262.144 seconds
b. How long does it take to take to download the file over a 1 megabit/second
modem?
T1M = 8 (1024) (1024) bits / 1x106 bits/sec = 8.38 seconds
c. Suppose data compression is applied to the text file. How much do the
transmission times in parts (a) and (b) change?
If we assume a maximum compression ratio of 1:6, then we have the following times
for the 32 kilobit and 1 megabit lines respectively:
T32k = 8 (1024) (1024) / (32000 x 6) = 43.69 sec
T1M = 8 (1024) (1024) / (1x106 x 6) = 1.4 sec
2. A scanner has a resolution of 600 x 600 pixels/square inch. How many bits are
produced by an 8-inch x 10-inch image if scanning uses 8 bits/pixel? 24 bits/pixel? [3
marks – 1 mark for pixels per picture, 1 marks each representation]
Solution:
The number of pixels is 600x600x8x10 = 28.8x106 pixels per picture.
With 8 bits/pixel representation, we have: 28.8x106 x 8 = 230.4 Mbits per picture.
With 24 bits/pixel representation, we have: 28.8x106 x 24 = 691.2 Mbits per picture.
6. Suppose a storage device has a capacity of 1 gigabyte. How many 1-minute songs can
the device hold using conventional CD format? using MP3 coding? [4 marks – 2 marks
each]
Solution:
A stereo CD signal has a bit rate of 1.4 megabits per second, or 84 megabits per
minute, which is approximately 10 megabytes per minute. Therefore a 1 gigabyte
storage will hold 1 gigabyte/10 megabyte = 100 songs.
An MP3 signal has a lower bit rate than a CD signal by about a factor of 14, so 1
gigabyte storage will hold about 1400 songs.
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8. How many HDTV channels can be transmitted simultaneously over the optical fiber
transmission systems in Table 3.3? [2 marks]
Solution:
Suppose that an optical fiber carries 1600 x 109 bps, and an HDTV channel is about
38 Mbps, then the fiber can carry about 1600000/38 = 40,000 HDTV channels.
60. Let g(x)=x3+x+1. Consider the information sequence 1001.
Solutions follow questions:
a. Find the codeword corresponding to the preceding information sequence.
Using polynomial arithmetic we obtain: [3 marks]
1011
1010
1001000
1011
01000
1011
00110
Codeword = 1001110
b. Suppose that the codeword has a transmission error in the first bit. What does the
receiver obtain when it does its error checking? [2 marks]
1011
0001
0001110
1011
101
CRC calculated by Rx = 101 Î error
62. Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111
00000000, 11110000 11110000, 11000000 11000000). Find the Internet checksum for
this code. [3 marks]
Solution:
b0 = 11111111 11111111 = 216 – 1 = 65535
b1 = 11111111 00000000 = 65280
b2 = 11110000 11110000 = 61680
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b3 = 11000000 11000000 = 49344
x = b0 + b1 + b2 + b3 modulo 65535 = 241839 modulo 65535 = 45234
b4 = −x modulo 65535 = 20301
So the Internet checksum = 01001111 01001101
63. Let g1(x) = x + 1 and let g2(x) = x3 + x2 + 1. Consider the information bits
(1,1,0,1,1,0).
a. Find the codeword corresponding to these information bits if g1(x) is used as the
generating polynomial. [2 marks]
11
100100
1101100
11
0011
11
0000
Codeword = 1101100
b. Find the codeword corresponding to these information bits if g2(x) is used as the
generating polynomial. [2 marks]
1101
100011
110110000
1101
01000
1101
1010
1101
111
Codeword = 110110111
c. Can g2(x) detect single errors? double errors? triple errors? If not, give an
example of an error pattern that cannot be detected. [2 marks – 0.5 each]
Single errors can be detected since g2(x) has more than one term. Double errors
cannot be detected even though g2(x) is primitive because the codeword length
exceeds 2n-k-1=7. An example of such undetectable error is 1000000010. Triple
errors cannot be detected since g2(x) has only three terms.
d. Find the codeword corresponding to these information bits if g(x) = g1(x) g2(x) is
used as the generating polynomial. Comment on the error-detecting capabilities
of g(x). [4 marks – 2 marks for the codeword and 2 for the comment]
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10111
111101
1101100000
10111
11000
10111
11110
10111
10010
10111
010100
10111
0011
Codeword = 1101100011
The new code can detect all single and all odd errors. It cannot detect double errors.
It can also detect all bursts of length n – k = 4 or less. All bursts of length 5 are
detected except for the burst that equals g(x). The fraction 1/2n-k = 1/16 of all bursts
of length greater than 5 are detectable.
67. Consider the m = 4 Hamming code.
a. What is n, and what is k for this code? [2 marks]
n = 2m − 1 = 15; k = n − m = 11
(15,11) Hamming code
b. Find parity check matrix for this code. [2 marks – 0.5 for each]
[000011111111000]
H= [111000011110100]
[011101100110010]
[101110101010001]
c. Give the set of linear equations for computing the check bits in terms of the
information bits. [2 marks – 0.5 for each]
b12 = b5 + b6 + b7 + b8 + b9 + b10 + b11
b13 = b1 + b2 + b3 + b8 + b9 + b10 + b11
b14 = b2 + b3 + b4 + b6 + b7 + b10 + b11
b15 = b1 + b3 + b4 + b5 + b7 + b9 + b11
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Communication Networks (2nd Edition)
Chapter 4 Solutions
Solutions to Chapter 4
1. A television transmission channel occupies a bandwidth of 6 MHz.
Solutions follow questions:
a.
How many two-way 30 kHz analog voice channels can be frequency-division multiplexed in a
single television channel?
6 × 10 / 30 × 10 = 200 channels
6
b.
3
How many two-way 200 kHz GSM channels can be frequency-division multiplexed in a
single television channel?
6 × 10 / 200 × 10 = 30 channels
6
c.
3
Discuss the tradeoffs involved in converting existing television channels to cellular telephony
channels?
The biggest advantage of using existing television channels to provide cellular telephony
channels is the very large bandwidth that they occupy. In theory, one could divide each television
channel into many cellular telephone channels, as shown in parts (a) and (b). The frequency
reuse aspect of cellular networks would multiply these channels many times. This would be at
the cost of television channels that cover a broad region and presumably provide service to a
large audience.
2. A cable sheath has an inner diameter of 2.5 cm.
Solutions follow questions:
a.
Estimate the number of wires that can be contained in the cable if the wire has a diameter of 5
mm.
Ignoring empty space between the wires:
25 2
)
2
= 25 wires
5
π ( )2
2
π(
b.
Estimate the diameter of a cable that holds 2700 wire pairs.
Ignoring empty space between the wires:
d
π ( )2
2
5 2
π( )
2
= 2x2700 wires Æ
d ≅ 368 mm
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Communication Networks (2nd Edition)
Chapter 4 Solutions
3. Suppose that a frequency band W Hz wide is divided into M channels of equal bandwidth.
a.
What bit rate is achievable in each channel? Assume all channels have the same SNR.
Each user uses W/M bandwidth. Using Shannon’s Channel Capacity formula:
Bit rate = (
b.
W
) log 2 (1 + SNR ) bps
M
What bit rate is available to each of M users if the entire frequency band is used as a single
channel and TDM is applied?
In this case, the total bit rate afforded by the W Hz is divided equally among all users:
Bit rate =
c.
W log 2 (1 + SNR )
bps
M
How does the comparison of (a) and (b) change if we suppose that FDM requires a guard
band between adjacent channels? Assume the guard band is 10% of the channel bandwidth.
Because of the guard band we expect that the scheme in (b) will be better since the bit rate in (a)
will be reduced. In (a), the bandwidth usable by each channel is 0.9W/M. Thus, we have:
Bit rate = (0.9
W
) log 2 (1 + SNR ) bps
M
4. In a cable television system (see Section 3.8.2), the frequency band from 5 MHz to 42 MHz is
allocated to upstream signals from the user to the network, and the band from 550 MHz to 750 MHz is
allocated for downstream signals from the network to the users.
Solutions follow questions:
a.
How many 2 MHz upstream channels can the system provide? What bit rate can each
channel support if a 16-point QAM constellation modem is used?
The system can provide:
(42 − 5) MHz
= 18 upstream channels.
Mhz
2 Channel
For bandpass channels, we have:
R = ( 2 ×10 )
6
b.
pulses
bits
∗4
= 8 Mbps
sec ond
pulse
How many 6 MHz downstream channels can the system provide? What bit rates can each
channel support if there is an option of 64-point or 256-point QAM modems?
Similarly, the system can provide:
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(750 − 550) MHz
= 33 upstream channels.
Mhz
6 Channel
R64-point = (6 ×10 )
6
pulses
bits
∗6
= 36 Mbps
sec ond
pulse
R256-point = (6 ×10 )
6
pulses
bits
∗8
= 48 Mbps
sec ond
pulse
5. Suppose a radio transmission system has a large band of available bandwidth, say 1 GHz, which is
to be used by a central office to transmit and receive from a large number of users. Compare the
following two approaches to organizing the system:
Solutions follow questions:
a.
A single TDM system.
The channel will provide a maximum bit rate, say b bps. Each user will use b/n bps where n is
the number of users. The number of users will be limited by the SNR on the channel, the time
guard band required so collisions do not occur, and the rate at which users can switch their
transmitters on and off.
b.
A hybrid TDM/FDM system in which the frequency band is divided into multiple channels
and time division multiplexing is used within each channel.
Although in theory, the total bit rate provided by the system is the same, the hybrid solution may
be practically better because, if we assume again n users, each user timeslot is at a lower bit rate
and for a longer duration. Therefore, TDM synchronization and switching speed problems are
less significant than in the pure TDM version. By leveraging both the time and frequency domain,
the technological requirements posed by either one are lessened.
6. Suppose an organization leases a T-1 line between two sites. Suppose that 32 kbps speech coding is
used instead of PCM. Explain how the T-1 line can be used to carry twice the number of calls.
Solution:
If a 32 kbps speech coding (such as Adaptive Differential Pulse Code Modulation (ADPCM)) is
used, the bit rate required for each user is reduced to half of that required for PCM (64 kbps).
The format in which a T-1 system carries information needs to be carried. For example a frame
could now carry 48 4-bit samples.
Aside: Chapter 12 explains how advanced speech coding schemes reduce the bit rate while
maintaining speech quality.
7.
A basic rate ISDN transmission system uses TDM. Frames are transmitted at a rate of 4000
frames/ second. Sketch a possible frame structure. Recall that basic rate ISDN provides two 64
kbps channels and one 16 kbps channel. Assume that one-fourth of the frame consists of overhead
bits.
Solution:
Assuming that the 16 kbps “D” channel is followed by the two 64 kbps “B” channels and then the
overhead (which is not necessarily the case), in the time domain, the frame could look as follows:
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Communication Networks (2nd Edition)
D channel
Chapter 4 Solutions
B channel
Overhead bits
B channel
t [ms]
0
0.021
0.104
0.1875
0.25
8. The T-1 carrier system uses a framing bit to identify the beginning of each frame. This is done by
alternating the value of the framing bit at each frame, assuming that no other bits can sustain an
alternating pattern indefinitely. Framing is done by examining each of 193 possible bit positions
successively until an alternating pattern of sufficient duration is detected. Assume that each
information bit takes a value of 0 or 1 independently and with equal probability.
Solutions follow questions:
a.
Consider an information bit position in the frame. Calculate the average number of times this
bit position needs to be observed before the alternating pattern is found to be violated.
Define the random variable X as the number of times the bit position needs to be observed before
the alternating pattern is violated. Assume the information bits are equally likely to be 0 or 1.
Without loss of generality assume first observed bit is zero.
P[X ≤ 1] = 0
P[X = 2] = ½
P[X = 3] = ¼
3
P[X = 4] = (½)
P[X = n] = (½)
E[X] =
=
since at least two observations are required
observe 0
observe 11
observe 10
(n -1)
∞
∞
i=2
i =2
¦ iP[ X = i]= ¦ ip
i −1
∞
d i
d ∞ i
d ∞ i
p =
p =
[¦ p − 1 − p ]
¦
dp i = 2
dp i = 0
i = 2 dp
=¦
d
1
1
1
− 1 − p] =
−1 =
−1
[
2
dp 1 − p
(1 − 1 2) 2
(1 − p )
=3
Note that two frame times elapse in the time required to make three observations.
b.
Now suppose that the frame synchronizer begins at a random bit position in the frame.
Suppose the synchronizer observes the given bit position until it observes a violation of the
alternating pattern. Calculate the average number of bits that elapse until the frame
synchronizer locks onto the framing bit.
On average, if the frame synchronizer starts at a random location, it will have to examine 193/2 =
96.5 incorrect bit locations before locking onto the framing bit. For each such incorrect bit
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Communication Networks (2nd Edition)
Chapter 4 Solutions
location, on average, a non-alternating pattern will be discovered on the third observation, at
which time the synchronizer will move one bit forward to the next bit.
Thus, the average number of bits elapsed until the synchronizer observes the framing bit is:
Average number of bits elapsed until frame bit found
= 96.5 x [(2 x 193) + 1] = 37345 bits
At this point, the synchronizer will begin observing the framing bit. It can’t be sure that it has
located the framing bit until an alternating pattern has been observed for a sufficient number of
bits, say n. The synchronizer can never be sure that it has locked onto the frame bit, so n must
be chosen sufficiently large so that the synchronizer is reasonably confident that it has indeed
found the frame bit.
9. The CEPT-1 carrier system uses a framing byte at the beginning of a frame.
a.
Suppose that all frames begin with the same byte pattern. What is the probability that this
pattern occurs elsewhere in the frame? Assume that each information bit takes a value of 0 or
1 independently and with equal probability.
The probability that random information matches the framing pattern is quite small:
8
p = (½) =1/256
b.
Consider an arbitrary information bit position in the frame. Calculate the average number of
times that the byte beginning in this bit position needs to be observed before it is found to not
be the framing byte.
The length N of a CEPT frame is 32 bytes, so N=256. The average number of frames M
observed before a location mismatches the framing pattern is:
M= 0(1− p)+ 1 p(1− p)+ 2 p 2 (1− p) + ... = p(1− p){1 + 2 p + 3 p 2 + ...
= p (1 − p )
1
p
(1 / 2)8
=
=
(1 − p) 2 1 − p 1 − (1 / 2)8
The first observation usually indicates that this is not the right position.
c.
Now suppose that the frame synchronizer begins at a random bit position in the frame.
Suppose the synchronizer observes the byte beginning in the given bit position until it
observes a violation of the alternating pattern. Calculate the average number of bits that
elapse until the frame synchronizer locks onto the framing byte.
The average number of bits observed until the beginning of frame is found is then:
N
28 1 / 28
28 2 − 1 / 28
N
+
=
) ≈ 256bits
( MN + 1) = (
1
)
(
2
2 1 − 1 / 28
2 1 − 1 / 28
10. Suppose a multiplexer has two input streams, each at a nominal rate of 1 Mbps. To accommodate
deviations from the nominal rate, the multiplexer transmits at a rate of 2.2 Mbps as follows. Each
group of 22 bits in the output of the multiplexer contains 18 positions that always carry information
bits, nine from each input. The remaining four positions consist of two flag bits and two data bits.
Each flag bit indicates whether the corresponding data bit carries user information or a stuff bit
because user information was not available at the input.
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a.
Chapter 4 Solutions
Suppose that the two input lines operate at exactly 1 Mbps. How frequently are the stuff bits
used?
In this case, the stuff bits are always used because the information bits alone only provide an
aggregate bit rate of 1.8 Mbps.
b.
How much does this multiplexer allow the input lines to deviate from their nominal rate?
This multiplexer provides either 9 or 10 bits for each stream per 22-bit frame. Thus, it allows
either of the two input streams to transmit as low as 0.9 Mbps and as high as 1.0 Mbps.
11. Calculate the number of voice channels that can be carried by an STS-1, STS-3, STS-12, STS-48,
and STS-192. Calculate the number of MPEG2 video channels that can be carried by these systems.
Solution:
A conventional voice channel is 64 kbps. An MPEG2 video signal can vary in bit rate. Assuming
a bit rate of 5 Mbps, the results are summarized in the table below:
STS-1
STS-3
STS-12
STS-48
STS-192
Bit rate of signal (Mbps) Voice Channels
51.84
810
155.52
2,430
622.08
9,720
2488.32
38,880
9953.28
155,520
MPEG2 Video
10
31
124
498
1,991
12. SONET allows positive or negative byte stuffing to take place at most once every four frames.
Calculate the minimum and maximum rates of the payload that can be carried within an STS-1 SPE.
Solution:
STS-1 rate = 51.84 Mbps
Payload rate = 50.112 Mbps
Frame = 125 µsec
Drifts = ( 8 bits ) / ( 4 x 125 µsec ) = 0.016 Mbps
Maximum payload rate = 50.112 + 0.016 = 50.128 Mbps
Minimum payload rate = 50.112 – 0.016 = 50.096 Mbps
13. Consider a SONET ring with four stations. Suppose that tributaries are established between each
pair of stations to produce a logical topology. Find the capacity required in each hop of the SONET
ring in the following three cases, assuming first that the ring is unidirectional and then that the ring is
bidirectional.
Note: The solution to this problem is currently being reviewed and may be revised.
Solutions follow questions:
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a.
Chapter 4 Solutions
The traffic between each pair of stations is one STS-1.
Unidirectional: STS-6
Bidirectional: STS-3
b.
Each station produces three STS-1s worth of traffic to the next station in the ring and no
traffic to other stations.
Unidirectional: STS-3
Bidirectional: STS-3
c.
Each station produces three STS-1’s worth of traffic to the furthest station along the ring and
no traffic to other stations.
Unidirectional: STS-6
Bidirectional: STS-6
14. Consider a set of 16 sites organized into a two-tier hierarchy of rings. At the lower tier a
bidirectional SONET ring connects four sites. At the higher tier a bidirectional SONET ring connects
the four lower-level SONET rings. Assume that each site generates traffic that requires an STS-3.
The high-tier ring must carry all inter-ring traffic from the lower tiers. The capacity required on the
high-tier ring is thus dependent on the volume of inter-ring traffic generated from the lower rings
and the distribution of such traffic. For this question, let us assume that the traffic generated is
distributed evenly throughout the high-tier network – that is, each low-tier ring sends an equal
amount of traffic to each of the other low-tier rings.
The lower-tier ring must accommodate 100% of the traffic sent from each of its sites. Since each
site will receive as much traffic as it will generate, each site will receive STS-3 of bandwidth.
Solutions follow questions:
a.
Discuss the bandwidth requirements that are possible if 80% of the traffic generated by each
site is destined to other sites in the same tier ring.
If 80% of the traffic is to stay within the same ring, then the required bandwidth on the higher tier
ring will be reduced as most traffic is confined locally to the lower tier rings.
In the lower tier rings the number of links to be traversed by the average message is between one
and two. Assuming that the traffic is distributed evenly on all of the links in the lower tier ring, the
bandwidth required on each link will be four times the average traffic between each two links,
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Communication Networks (2nd Edition)
Chapter 4 Solutions
which is 4x 1/3 x 80% STS-3 or 16/15 STS-3. Adding the other 20% of the traffic which is
destined to the higher tier link and has to traverse some of the lower tier links in both directions,
the average link capacity in the lower tier links should be 16/15 STS-3 + 2 x 20% STS-3 or 22/15
STS-3, which is almost 1.5 STS-3 on average.
On the higher tier ring, the traffic from each subring will be routed equally to each other subring.
Thus the bandwidth required on each link will be four times the average traffic between two
subrings or 4 x 1/3 of the total traffic from each subring. Because each ring produces 4×0.2 ×
STS-3 amount of traffic, the bandwidth required on each link of the higher tier ring is on the order
of 4/3 × 0.8 × STS-3 ≈ STS-3.
b.
Discuss the bandwidth requirements that are possible if 80% of the traffic generated by each
site is destined to sites in other rings.
If 20% of the traffic is to stay within the same ring, then the required bandwidth on the higher tier
ring will be increased. The required bandwidth on the lower tier rings will not change significantly,
as each source still produces and receives STS-3 of traffic.
The higher tier ring will now support 4 times more traffic. Thus, the bandwidth requirement is
around STS-12.
15. Compare the efficiency of BLSR and UPSR rings in the following two cases:
a.
All traffic originating at the nodes in the ring are destined for a given central node.
Considering that the total capacity of working and protection lines in both BLSR and
UPSR is the same, the total bandwidth utilized in both BLSR and UPSR in this scenario is
the same. The difference is the distribution of the traffic and unused capacity across the
ring in the two cases.
As an example we consider the ring shown in the figure below, consisting of four nodes.
We first consider the UPSR scenario. The last hop before the central node is the
bottleneck in this case limiting the traffic in the other hops on the working ring shown with
solid lines. Assuming equal distribution of bandwidth among the nodes, if the last hop is
fully utilized the hop prior to the last hop will be utilized up to 2/3 of its capacity and the
one before that will be utilized up to 1/3 of its capacity. The remaining hop will not be
utilized at all. The lines on the protection path will be utilized similarly but in the opposite
order, shown with dotted lines. As a result the sum of working and protection lines
together will be utilized only to half of the total capacity all across the ring.
UPSR
BLSR
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Chapter 4 Solutions
Now we consider the BLSR scenario. In this case the working traffic can be sent in both
directions but the protection traffic will be sent in the opposite direction and as a result as
seen in the figure the total of working and protection traffic all across the ring will be the
same as the other case as well as the unused capacity of the links.
b.
Each node originates an equal amount of traffic to all other nodes.
In both BLSR and UPSR lines can be shared by traffic from other nodes. The more nodes
share the same line the less the line can be utilized by the node attached to the line. In the
given scenario each line is shared by the traffic from twice the number of nodes in UPSR
compared to the BLSR. As a result BLSR is more efficient in this case.
Working
ring in
BLSR
Working
ring in
UPSR
16. Consider the operation of the dual gateways for interconnecting two bidirectional SONET rings
shown in Figure 4.29. The primary gateway transmits the desired signal to the other ring and
simultaneously transmits the signal to the secondary gateway which also routes the signal across the
ring and then to the primary gateway. A service selector switch at the primary gateway selects
between the primary and secondary signals. Explain how this setup recovers from failures in the link
between the primary gateways.
Solution:
Two logical connections are maintained between the primary gateways. The first is a direct
physical connection via the primary gateway link. The second is connected indirect via the
secondary gateway as described in the question.
This redundancy provides a backup in case of a link failure. If the primary gateway link fails, the
service selector switch can select signals for the alternate indirect path until the primary gateway
link is restored.
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Chapter 4 Solutions
17. Consider the synchronous multiplexing in Figure 4.15. Explain how the pointers in the outgoing
STS-1 signals are determined.
Solution:
The section and line overhead of each incoming STS-1 signal is terminated and its payload is
mapped into a new STS-1 frame that is synchronous to a local clock. The outgoing pointers are
all adjusted according to a common clock such that all the outgoing STS-1 frames are
synchronized with respect to each other.
18. Draw a sketch to explain the relationship between a virtual tributary and the synchronous payload
envelope. Show how 28 T-1 signals can be carried in an STS-1.
Solution:
Seven tributaries are multiplexed in stream. Each tributary carries four T-1 channels and
occupies twelve columns of the SPE, as shown below:
87 Columns
T1
T2
T3
T4
T5
T6
T7
9 rows
t
The first column is path overhead. The next 84 columns are the seven tributaries and the last two
columns are unused.
19. Do a web search for DWDM equipment available from telecommunications equipment vendors.
What specifications do you see in terms of number of channels, spacing between channels, bit rates
and formats of the signals that can be carried?
Solution:
Some specs for WDM and DWDM multiplexers:
Wavelength: 632- 820 nm, 980-1064 nm, 980-1480 nm, 980-1550 nm, 1064-1310 nm,
1310-1550 nm, 1310-1625 nm, 1480-1550 nm, 1550-1625 nm, etc.
Channel Spacing: 50 GHz (0.1 nm spacing between channels), 100 GHz (0.2 nm spacing
between channels), 200 GHz (0.4 nm spacing between channels), etc.
Per channel bit rates: 100 Mbps, 1.2 Gbps, 2.4 Gbps, 10 Gbps, etc.
Number of Channels: 4, 8, 16, 32, 80, 160
20. Consider WDM systems with 100, 200, and 400 wavelengths operating at the 1550 nm region and
each carrying an STS-48 signal.
Solutions follow questions:
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a.
Chapter 4 Solutions
How close do these systems come to using the available bandwidth in the 1550 nm range?
One STS-48 signal carries 2488.32 Mbps of traffic. Thus, the aggregate bit rate of each of these
systems is:
System
Bit rate
100 wavelengths
248.832 Gbps
200 wavelengths
400 wavelengths
497.664 Gbps
995.328 Gbps
The available bandwidth of this window is 12.49 THz. Thus, none of these systems even comes
close to using all of the available bandwidth.
b.
How many telephone calls can be carried by each of these systems? How many MPEG2
television signals?
The number of telephone calls (64 Kbps) and MPEG2 video signals (5 Mbps) that can be carried
each system is shown below:
System
100 wavelengths
200 wavelengths
400 wavelengths
Telephone Calls
6
3.888 × 10
6
7.776 × 10
6
15.552 × 10
MPEG2 Video Signals
49 766
99 532
199 065
Although, none of the systems comes close to using all of the available bandwidth, they can still
carry huge amounts of traffic. This shows how WDM provides the potential for huge amounts of
bandwidth and radically expands the bandwidth of existing optical links.
21. Consider the SONET fault protection schemes described earlier in the chapter. Explain whether
these schemes can be used with WDM rings.
Solution:
The fault protection schemes described in problem 15 are applicable to any ring network
structure. To use these schemes in WDM, the exact same procedures can be carried out except
that in the case of WDM, instead of switching streams, wavelengths are switched. Furthermore,
the switching can be done in optical domain by all optical protection switches that detect the
signals and switch to protection signal instantaneously. Optical protection switching can be faster
and more efficient.
22. Calculate the spacing between the WDM component signals in Figure 4.8. What is the spacing in
hertz and how does it compare to the bandwidth of each component signal?
Solution:
In Figure 4.8 there are 16 wavelengths distributed equally over the spectrum from 1545 nm to
1557 nm. Thus the spacing between each wavelength is:
1557 − 1545
= 0.75nm
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The spacing in Hertz is:
Frequency Bandwidth =
(3 ∗10 8 )(0.75 ∗ 10 −9 )
= 93.65GHz
(1550 ∗ 10 −9 ) 2
Thus, the maximum achievable bit rate that each wavelength can achieve using on-off modulation
is 2 × 93.65 = 187 Gbps. Each signal carries 2.5 Gbps of traffic, so we see that because of the
relatively large guard bands needed, WDM wavelengths have a very large spacing compared to
their width.
23. How does WDM technology affect the hierarchical SONET ring topology in Figure 4.29? In other
words, what are the consequences of a single fiber providing large number of high-bandwidth
channels?
Solution:
WDM allows for much more aggregate bandwidth on each fiber. Thus, in the context of the
hierarchical SONET ring topology, the high-tier ring can use WDM in order to increase its
bandwidth and the number of channels it can support and, in turn, the number of lower-tier rings
that it can service. WDM allows it to achieve this increased service ability without the need to lay
down more physical fibers over its links. On the other hand, the use SONET ring topology mean
that each wavelength is associated with a separate SONET ADM, and this creates a problem in
terms of interconnecting traffic across rings.
24. WDM and SONET can be used to create various logical topologies over a given physical
topology. Discuss how WDM and SONET differ and explain what impact these differences have in
the way logical topologies can be defined.
Solution:
Both WDM and SONET define logical topologies by setting up semi-permanent paths between
nodes in the network. WDM creates these paths by assigning lightpaths between nodes. WDM
without wavelength conversion has an inherent restriction imposed on the logical topologies that it
can create, because no two lightpaths that share the same link can use the same wavelength. In
SONET, since tributaries are electrically defined, the only restriction imposed on the topologies is
the total bandwidth utilized on each link.
25. Compare the operation of a multiplexer, an add-drop multiplexer, a switch, and a digital crossconnect.
Solution:
A multiplexer is a 1:N device. It takes N separate signals on N different inputs and combines
them into one higher rate signal on one output port.
An add-drop multiplexer takes in N signals on one input port and replaces one of them with a new
signal from a separate input port. The new aggregate signal is routed to an output port and the
signal that was replaced is “dropped” to a separate output port.
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A switch takes in N inputs and routes them to N different outputs. Frequently each input to a
switch contains a number of multiplexed connections. These connections can be demultiplexed
and routed to different output ports where they are aggregated prior to exiting the switch.
Generally, switches are configured using signaling that establishes paths across the network.
A digital cross-connect is similar to a switch except that it is semi-permanent, usually configured
by network operators rather than signaling processes. Their configuration is also done on a
larger time scale (days or weeks). Digital cross-connects provide a basic network topology on
which routing can be applied.
26. Consider a crossbar switch with n inputs and k outputs.
Solutions follow questions:
a.
Explain why the switch is called a concentrator when n > k. Under what traffic conditions is
this switch appropriate?
This switch is a concentrator because traffic comes in on n lines and is concentrated onto k lines.
Traffic on the output lines is higher than that of the input lines by a factor of n/k on average. The
switch has inherent multiplexing functionality.
b.
Explain why the switch is called an expander when n > k. Under what traffic conditions is
this switch appropriate?
This switch is an expander because the number of lines is expanded from n to k. This switch is
appropriate when the outgoing lines carry less traffic than the incoming lines. One such example
is at the egress of a backbone network to an access network. The switch has inherent
demultiplexing functionality.
c.
Suppose an N x N switch consists of three stages: an N x k concentration stage; a k x k
crossbar stage; and a k x N expansion stage. Under what conditions is this arrangement
appropriate?
2
The number of crosspoints in an N x N crossbar switch is N . If the traffic load on the inputs is
relatively low, this configuration may save on hardware costs. The number of crosspoints needed
in the three-stage switch described is:
2
k + 2kN
2
2
Depending on the values of N and k, k + 2kN may be less than N , especially if N is much
greater than k.
d.
When does the three-stage switch in part (c) fail to provide a connection between an idle input
and an idle output line?
Assuming unicast traffic, if more than k inputs require a path to one of the third stage’s n outputs
at the same time, blocking will occur in the first stage, since it only has k outputs.
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27. Consider the multistage switch in Figure 4.35 with N = 16, n = 4, k = 2.
Solutions follow questions:
a.
What is the maximum number of connections that can be supported at any given time?
Repeat for k = 4 and k = 10.
For N = 16, n = 4 and k = 2, we have the following switch architecture:
4x2
2x4
4x4
4x2
2x4
4x2
2x4
4x4
4x2
2x4
Thus, the second stage is the bottleneck, and blocking can occur in the first stage. Thus, eight
connections can be supported at a time.
If k = 4, then blocking will occur if we are not allowed to rearrange connections. It can be shown
that in this case blocking can be avoided if we are allowed to rearrange the connection pattern
every time a new connection request is made.
If k = 10, then there are ten 4 x 4 switches in the second stage. Since there are only 16 inputs
and 16 outputs, the switch can accommodate any set of connections without blocking.
b.
For a given set of input-output pairs, is there more than one way to arrange the connections
over the multistage switch?
As shown in the picture in part (a), it is clear that each input-output pair can be connected through
any one of the k second-stage switches. Thus, there are k ways to arrange the connections over
a multi-stage switch.
28. In the multistage switch in Figure 4.35, an input line is busy 10% of the time.
Solutions follow questions:
a.
Estimate the percent of time p that a line between the first and second stage is busy.
Although, the busyness of each output line of the first stage depends on the switch
control, on average, we can estimate that each of the output lines will have n/k the amount
of traffic as the input lines. Thus
p = (n/k) 10%
b.
How is p affected by n and k?
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p is directly proportional to the ratio of n to k.
c.
How does this p affect the blocking performance of the intermediate crossbar switch?
Blocking occurs at the second switches when two of its inputs are to be routed to the same
output. Although the control and routing algorithm of the switch can minimize the chance of this
occurring, the greater the value of p, the greater the chance that such blocking will occur.
d.
Supposing that the blocking probability of the intermediate crossbar is small, what is the
proportion of time p′ that a line between the second and third stage is busy?
Assuming that the blocking probability of the second stage switches is small, the percentage of
time that the output lines on a switch in the second stage are busy is equal to the percentage of
time that its input lines are busy. This is the case because the second stage switches are
(N/n)x(N/n) components. Thus,
p′ = p = (n/k) 10%
e. For a given input and output line, what is the probability that none of the N/n paths between the input
and output lines are available?
In this question N/n should be k, since there are k possible paths from each input to each output
(one different path possible through each second-stage switch). Assuming independence of all
lines, which is not the case generally, for any one of the k paths between a given input and the
output to be unavailable, one of the 2 jumps must be unavailable.
P[path not available] = 1-P[path is available] = 1-(1-p’)(1-p’) ≈ 2p
For none of the paths to be available, all k must be busy. Since we assumed that all are
independent of each other,
P[none of the paths available] = (2p)
k
29. Consider the multistage switch in Figure 4.35 with N = 32. Compare the number of crosspoints
required by a nonblocking switch with n = 16, n = 8, n = 4, and n = 2.
Solution:
For any switch to be non-blocking, we require knb = 2n – 1. The total number of crosspoints is
2
2Nk + k(N/n) . The resulting number of crosspoints necessary for different values of n is shown
below.
N
32
32
32
32
n
16
8
4
2
k
31
15
7
3
Number of Crosspoints needed
2108
1200
896
960
For a one-stage N x N switch with n = 32, we would require 1032 crosspoints. Thus we see that,
just as was noted in question 26c, for k much less than n, multistage switches can provide good
hardware economy while remaining non-blocking.
30. A multicast connection involves transmitting information from one source user to several
destination users.
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Solutions follow questions:
a.
Explain how a multicast connection may be implemented in a crossbar switch.
In a crossbar switch, multicasting can be implemented by copying the multicast input to multiple
outputs.
b.
How does the presence of multicast connections affect blocking performance of a crossbar
switch? Are unicast calls adversely affected?
Unlike unicast switches, flow in multicast switches is not conserved. Generally, if multicasting is
implemented in a switch, the total traffic flow out of the switch is more than the flow into the
switch. Thus the probability of blocking is increased because the traffic inside the switch is
higher.
Unicast traffic is also affected by this increased flow because fewer resources in the switch are
available.
c.
Explain how multicast connections should be implemented in a multistage switch. Should the
multicast branching be done as soon as possible or as late as possible?
In a multistage switch with a given multicast traffic profile, each branching procedure represents
an increase in traffic flow beyond the branch point. Thus, to keep traffic load in the switch to a
minimum, branching should be done as late as possible.
31. Do a web search for crosspoint switching chips. What specifications do you find for number of
ports, bit rate/port? Are dualcasting and multicasting supported? Do the crosspoints need to be
changed together or can they be changed independently? What considerations determine how fast the
crosspoints can be changed?
Solution:
Some specs for Crosspoint fabrics chips:
Switch size: 2x2, 4x4, 17x17, 34x34, 36x37, 68x69, 64x64, 72x72, 144x144
Port speed: 1.6 Gbps, 2.5 Gbps, 3.2 Gbps, 3.6 Gbps
Dualcasting and multicasting is usually supported by crosspoint chips, as well as masking
particular ports, and also programming a single port or a subset of ports independently.
The speed of changes depend on the clock recovery and signal synchronization at the
port interfaces (receivers and transmitters at input and output ports of the crosspoint
fabric), and also the speed of programming the I/O Table registers.
32. What is the delay incurred in traversing a TSI switch?
Solution:
The arriving frame is written onto the switch register and after the frame is completely written in, it
needs to be read out in permuted order.
33. Explain how the TSI method can be used to build a time-division multiplexer that takes four T-1
lines and combines them into a single time-division multiplexed signal. Be specific in terms of the
number of registers required and the speeds of the lines involved.
Solution:
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Each input line carries T-1 traffic. The input frame size is n = 24 slots. Each frame that is output
is of length 4n = 96 slots. After a full set of 96 slots is written into the 4 registers, the slots are
read out in the outgoing line.
1.544 Mbps
AA
1.544 Mbps
BA
1.544 Mbps
CA
1.544 Mbps
DA
24
slots
24
slots
AA BA CA DA
6.176 Mbps
24
slots
24
slots
34. Suppose that the TDM frame structure is changed so that each frame carries two PCM samples.
Does this affect the maximum number of channels that can be supported using TSI switching?
Note: The solution to this problem is currently being reviewed and may be revised.
Solution:
Yes. Now each slot requires two memory reads and two memory writes. Thus, the maximum
number of channels that can be supported is:
Max =
125µs
4 × mem
In other words, one half as many slots per frame can be supported as compared to when each
slot requires one PCM sample.
35. Examine the time-space-time circuit-switch architecture and explain the elements that lead to
greater compactness, that is, smaller physical size in the resulting switching system.
Solution:
In a space switch, n inputs are switched simultaneously in a switching time of T seconds. The
space-time circuit switch switches each of the n inputs in succession with a switching time of T/n
each, for a total switching time of T.
By switching the inputs one at a time instead of simultaneously, the switching matrix needed at
any given time is reduced in size accordingly and greater hardware compactness is achieved.
The implicit tradeoff, however, is that higher line and switch speeds are needed.
36. Consider the three-stage switch in problem 4.26c. Explain why a space-time-space
implementation of this switch makes sense. Identify the factors that limit the size of the switches that
can be built using this approach.
Solution:
A space-time-space implementation would have the middle stage replaced by a TSI switch stage,
and thus reduce hardware costs. The maximum number of slots in a TSI switch, say M, is limited
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by the length of the memory cycle. This limitation means that N/n ≤ M, which in turn places a
limitation on N, the number of inputs to the switch (since n has a maximum that is determined by
technological constraints). The limitation on the number of slots M suggests that the approach is
suitable when concentration can be used in the input stage because of light loading in the input
lines.
37. Consider n digital telephones interconnected by a unidirectional ring. Suppose that transmissions
in the ring are organized into frames with slots that can hold one PCM sample.
Solutions follow questions:
a.
Suppose each telephone has designated slot numbers into which it inserts its PCM sample in
the outgoing direction and from which it extracts its received PCM sample from the incoming
direction. Explain how a TSI system can be used to provide the required connections in this
system.
At some point in the ring, TSI is carried out to interchange the information in the time slots
between all pairs of connected users
b.
Explain how the TSI system can be eliminated if the pairs of users are allowed to share a time
slot?
If all user pairs share a time slot, for any connected pair of users we can assign the same timeslot to both. To communicate, each of the users inserts outgoing PCM samples on the slot and
extracts incoming PCM samples from the same slot.
c.
How would connections be established in parts (a) and (b)?
In the system described in part (a) the TSI switch monitors all of the slots continuously for
connection request made by the users. If user A requests a connection to user B, the central
switch controller makes the connection if the slot corresponding to user B is available.
In the system described in part (b) a mechanism is needed to coordinate the sharing of slots
between all the pairs of users. The most efficient mechanism would have half as many timeslots
as users. The central switching mechanism would assign each pair of users a timeslot to
communicate over. Call setup would require all users to communicate with the central switch on
a setup slot that would have to be accessed via a MAC protocol.
An alternative simple distributed mechanism would involve each user monitoring their own timeslot. Assuming that the users have time slot tunability, a call could be initiated by sending a
sample on the intended receivers time slot (provided it is free). Communication would occur on
the receiver’s time slot in this scenario. The drawback of this mechanism is that no more than half
of the available bandwidth would ever be used.
38. Consider the application of a crossbar structure for switching optical signals.
a.
Which functions are the crosspoints required to implement?
The crosspoints must be able to open, close, hold open, and hold closed connections.
b.
Consider a 2 x 2 crossbar switch and suppose that the switch connection pattern is (1 → 1, 2
→ 2) for T seconds and (1 → 2, 2 → 1) for T seconds. Suppose it takes τ seconds to change
between connection patterns, so the incoming optical signals must have guardbands to allow
for this. Calculate the relationship between the bit rate R of the information in the optical
signals, the number of bits in each “frame”, and the values T and τ? For R in the range from 1
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gigabit/second to 1 terabit/second and τ in the range of 1 microsecond to 1 millisecond, find
values of T that yield 50% efficiency in the use of the transmission capacity.
The transition time is τ seconds, and the hold time is T seconds. The optical transmission rate is
R bps and the actual information bit rate is Reffective, and letting n be the number of bits per frame
(including overhead), we have:
Reffective = (total bit rate of connection)(proportion of useful bits)
=R∗(
T
n
)=
T +τ
T +τ
The value of T that yields 50% efficiency is T = τ.
For T = τ = 1 microsecond, we have n = RT which gives frame size of n = 1000 bits at 1 Gbps
and n = 1000000 bits at 1 Tbps. For T = τ = 1 millisecond, the frame sizes are 1000 times longer.
39. Suppose an optical signal contains n wavelength-division multiplexed signals at wavelengths λ1,
λ2, …, λn. Consider the multistage switch structure in Figure 4.35 and suppose that the first stage
switches consist of an element that splits the incoming optical signal into n separate optical signals
each at one of the incoming wavelengths. The jth such signal is routed to jth crossbar switch in the
middle stage. Explain how the resulting switch structure can be used to provide a rearrangeable optical
switch.
Solution:
In the resulting architecture, the second-stage inputs to any given switching element have the
same wavelength. Thus, each single wavelength signal input to the second stage can be directed
to any one of the third stage elements. Each third stage element corresponds to a single output
port. It is assumed that the third stage consists of elements that recombine the n incoming
signals, each with a different wavelength, into a WDM signal. The operation of this switch is very
similar to that of the switch in figure 4.35 However, it differs in that there is only one possible
route for each input-output-wavelength combination.
40. Consider the equipment involved in providing a call between two telephone sets.
Solutions follow questions:
a.
Sketch a diagram showing the various equipment and facilities between the originating
telephone through a single telephone switch and on to the destination telephone. Suppose first
that the local loop carries analog voice; then suppose it carries digital voice.
In an analog connection, the telephone set is connected by a pair of wires to the telephone office.
The wires are terminated in a line card that digitizes the voice signal and transfers the digital
signal to a digital switching system. The digital signal travels along a fixed path from the
telephone switch in the originating office to the telephone switch in the destination office. The two
or more offices are interconnected by digital transmission lines.
If the telephone set were to generate digital voice then a digital signal would be transmitted along
the local loop that connects the telephone to the central office. The operation of the line card
would differ, as the arriving signal would already be in digital form.
b.
Repeat part (a) in the case where the two telephone calls are in different LATAs.
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If the two telephones are in different LATAs then the digital voice signal must traverse
tandem switches that connect the two LATAs.
c.
In parts (a) and (b) identify the points at which a call request during setup can be blocked
because resources are unavailable.
Call requests can be blocked when there are insufficient resources to complete the call. These
resources can include time slots and crosspoints in the switches and time slots in the digital lines.
Blocking can also result from unavailability of signaling processing equipment.
41. Suppose that an Internet service provider has a pool of modems located in a telephone office and
that a T-1 digital leased line is used to connect to the ISP’s office. Explain how the 56K modem (that
was discussed in Chapter 3) can be used to provide 56 kbps transfer rate from the ISP to the user.
Sketch a diagram showing the various equipment and facilities involved.
Solution:
The ISP office is connected to the user’s local central office, which is in turn connected to the
user. The ISP can use the T-1 line to send digital information at a rate of 56 kbps into and across
the telephone network to the line card that connects to the user. The line card can then deliver
information at 56 kbps to the user. Recall that the user cannot send at 56 kbps because of
limitations on the available SNR.
CO
Modem
ISP office
T-1 line
Twisted Pair
User’s
comp
Modem bank
42. Why does a conventional telephone still work when the electrical power is out?
Solution:
The telephone company supplies each of its telephone lines with power at the central office. This
power is stored in the form of wet batteries that can alternately be charged by a backup battery in
the event of a power failure at the central office. These huge batteries occupy entire floors in
telephone offices.
43. In Figure 4.44b, how does the network know which inter-exchange carrier is to be used to route a
long distance call?
Solution:
The inter-exchange carrier that handles the long-distance calls for a specific user is specified a
priori by the user’s service contract. The user “signs-up” with one long distance plan over
another, and it is the responsibility of the local service provider to route the call over the
appropriate inter-exchange carriers.
44. ADSL was designed to provide high-speed digital access using existing telephone facilities.
a.
Explain how ADSL is deployed in the local loop.
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An ADSL modem is required at each end of the user’s twisted pair connection. Since ADSL
coexists with regular phone service, it uses higher frequencies for data transmission. Bandwidthlimiting elements, such as loading coils, must be removed. ADSL can only be offered within
certain distances of the central office.
b.
What happens after the twisted pairs enter the telephone office?
At the telephone office, the voice and data are separated/filtered and the demultiplexed signals
are routed accordingly (for example, voice may be routed over the phone network and data over
the Internet).
c.
Can ADSL and ISDN services be provided together? Explain why or why not.
No, ADSL and ISDN use overlapping parts of the signal spectrum so they cannot operate
simultaneously.
45. In this problem we compare the local loop topology of the telephone network with the coaxial
cable topology of cable television networks (discussed in Chapter 3).
a.
Explain how telephone service may be provided using the cable television network.
Since the television network is inherently a broadcast, shared medium, a method such as the
cable modem method described in chapter 3 would be needed to accommodate the traffic of
predominantly point-to-point bidirectional telephone service. The mechanism used would include
a MAC protocol of some form to provide a fair method of partitioning the total available bandwidth
among callers.
b.
Explain how cable television service may be provided using the local loop.
It would be very difficult to provide cable television on the local loop network due to the huge
bandwidth currently provided by the cable broadcast network. The only effective means of
providing full cable service over the local loop would likely involve sending a single channel to
each user. Changing channels would require an upstream request to the local switch, which
would then replace the originally transmitted channel with the requested one. By moving the
pedestal that connects to the home, it may be possible to provide sufficiently high speeds to offer
several simultaneous channels to a home.
c.
Compare both topologies in terms of providing Internet access service.
Both these networks currently provide Internet access. Cable modems have the advantage of
providing huge peak bandwidth via that large bandwidth (up to 30 Mbps) of the cable network.
However, as has been discovered by many subscribers to this service, since the cable network is
a shared broadcast medium, the bandwidth provided does not scale well to many users. During
peak hours, the total bandwidth of the network is divided among all of the users connected to the
final branching point of the network. This leads to noticeable and significant decrease in service.
Access to the Internet via the phone network using DSL modems has the advantage of providing
a dedicated access line to its users. As a result, each user’s service access rate is independent
of the number of users. The tradeoff is the lower bandwidths available. Currently ADSL typically
provides bandwidth of up to 1 Mbps.
46. The local loop was described as having a star topology in the feeder plant and a star topology in
the distribution plant.
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Compare the star-star topology with a star-ring topology and a ring-ring topology. Explain
how information flows in these topologies and consider issues such as efficiency in use of
bandwidth and robustness with respect to faults.
Star topologies generally provide efficient use of bandwidth because their hierarchical routing
structure provides near shortest path routing. However, in situations with low load, ring networks
are more efficient. When a user is not transmitting any information in a star topology, the
bandwidth of the line to the user is wasted. In a ring topology, since the bandwidth is shared,
active users can make use of the extra available bandwidth provided by the inactive users.
In general, pure star topologies provide no mechanism for fault line recovery because there is not
backup path. Thus, this architecture should be used if the connections are very reliable.
The three topologies under examination are shown below.
The star-star topology is efficient when the load on each of the lines is high. If one of the lines on
a high tier star fails, service for many users is lost. Thus, such architecture is only good if the
lines are very reliable.
The star-ring topology is good when the traffic produced by each user is relatively sporadic, but
the traffic in the “star” links is high. This situation could arise if the bandwidth provided in “star”
links is limited in capacity. It still has the same vulnerability as the star-star network in term of
large scale loss of service if one of the high-tier links fails.
The ring-ring topology is very robust to faults because every path can be backed up by forming
the backup path around the ring in the other direction. In every link in the network, bandwidth is
shared and therefore used efficiently.
b.
What role would SONET transmission systems play in the above topologies?
SONET transmission systems could be used in the architectures described. It would be
particularly useful in the rings to provide OAM functions and fault protection.
47. Suppose that the local loop is upgraded so that optical fiber connects the central office to the
pedestal and twisted pair of length at most 1000 feet connects the user to the pedestal.
a.
What bandwidth can be provided to the user?
Using ADSL a bandwidth of 8 Mbps can be provided. Since the twisted pair length is only 1000
feet, VDSL may be used. In this case, a bandwidth of 55 Mbps can be provided.
b.
What bit rate does the optical fiber have to carry if each pedestal handles 500 users?
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If ADSL is used, and each fiber handles 500 users, the total bandwidth on the fiber must be at
least 500 × 8 Mbps = 4 Gbps. If VDSL is used, the bandwidth increases to 55 Mbps x 500 = 27.5
Gbps.
c.
How can SONET equipment be used in this setting?
SONET can be used at the physical layer in the optical fiber so that the multiplexing of the 500
lines is simplified using tributaries. This functionality could be extended to the pedestal to provide
OAM control of the large aggregate bandwidth connecting it to the central office.
48. Let’s consider an approach for providing fiber-to-the-home connectivity from the central office to
the user. The telephone conversations of users are time-division multiplexed at the telephone office
and broadcast over a “passive optical network” that operates as follows. The TDM signal is broadcast
on a fiber up to a “passive optical splitter” which transfers the optical signal to N optical fibers that are
connected to N users. Each user receives the entire TDM signal and retrieves its own signal.
Solutions follow questions:
a.
Trace the flow of PCM samples to and from the user. What bit rates are required if N=10,
100? Compare these to the bit rate that is available.
Because a passive splitter is used, the fiber must have enough bandwidth to support N
connections. Thus,
N = 10 Æ 640 kbps minimum capacity
N = 100 Æ 6.4 Mbps
Currently bit rates of 2.5 Gbps on a single wavelength fiber is common and 10 Gbps is available.
b.
Discuss how Internet access service might be provided using this approach.
Internet access could be very simply provided by replacing the PCM samples with Internet traffic.
c.
Discuss how cable television service might be provided using this approach.
Since MPEG2 television signals are 2-5 Mbps, on the order of 1000 channels could be provided
by this network to each user. Each TDM slot could correspond to one channel. The users could
change channels by resynchronizing their receivers to the appropriate time slot.
d.
What role could WDM transmission play in the connection between the central office and the
optical splitter? in the connection all the way to the user?
If WDM is used, the aggregate bandwidth of the optical fiber would be greatly increased.
Because the architecture is a passive broadcast system, if WDM were used between the central
office and the passive splitter, it would also be on the fiber to the user.
The additional bandwidth might still be leveraged in a number of ways. The additional
wavelengths provide an additional degree of multiplexing flexibility. For example, the users could
tune into one wavelength at a time (with tuning done on a relatively large time scale) and each
wavelength could carry a “cable package“ that includes a subset of the provided channels.
Although WDM provides increased bandwidth, a WDM receiver typically represents significantly
higher hardware costs. If the huge amounts of available bandwidth were used to carry
heterogeneous traffic, the hardware costs required would be offset by the benefits. The resulting
network could carry television, telephone, and Internet traffic and, in effect, replace all of the
currently existing access networks.
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Chapter 4 Solutions
49. Explain where the following fit in the OSI reference model:
Solutions follow questions:
a.
A 4 kHz analog connection across the telephone network.
Physical Layer: the actual 4 kHz analog signal exists only in the physical layer of the OSI
reference model.
b.
A 33.6 kbps modem connection across the telephone network.
Data-Link Layer: a 33.6 kbps modem uses framing, flow-control, and error correction to connect a
user to the switch.
c.
A 64 kbps digital connection across the telephone network.
Physical Layer: the digital link across the network is controlled by many higher layer functions, but
the 64 kbps signal that carries user information is analogous to the 4 kHz signal used over the
twisted pair that runs to the user’s premises.
50. For the following examples of connections, explain the signaling events that take place inside the
network as a connection is set up and released. Identify the equipment and facilities involved in each
phase of the connection.
Solutions follow questions:
a.
A voice call in the telephone network.
SSP directs the signaling information to signaling network. SCP sends required
commands to individual switches to setup the path. The same procedure takes place to
disconnect the call.
b.
A frame relay connection.
Frame relay is a connection oriented network. Once the request for a connection is made
all the nodes are programmed to direct the frames to appropriate route.
c.
A SONET connection in metropolitan network as shown in Figure 4.29.
Connection setup requires finding available tributary or STS-n in local rings as well as the
metro ring, and regional ring if required. The path is setup by programming the nodes
across the path to use the selected tributary or STS-n streams for the connection. Once
this is done the transmission can start. To disconnect all of the assigned streams are
released.
d.
An optical connection across the continent.
To setup an optical connection across the continent an available wavelength is to be
found for every piece of the route from source to destination. Once the wavelength or set
of wavelengths is found at each node the switch is programmed or manually set up to
direct the signal in the right output according to the path information. To disconnect the
wavelengths are set to be free.
51. Sketch the sequence of events that take place in the setting up of a credit-card call over the
intelligent network. Identify the equipment involved in each phase.
Solution:
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Before a connection is made, a recorded message requests the credit card number for the call.
The user inputs the digits and the tones are converted to numbers and the message sent to a
database in the appropriate SCP. The credit card number is verified and may be used to check
for the authorization of funds. The results of the query are sent to the SCP. The appropriate
recorded message is sent back to the user. Based on the query, the call is accepted and a
connection made, or the call is denied.
52. Explain how the intelligent network can provide the following services:
Before a connection is made every call has its signaling information in the control plane routed
through an SCP that stores the destination phone's profile (that is, identifies the services that
have been subscribed).
a.
Caller identification – A display on your phone gives the telephone number or name of the
incoming caller.
When the ring signal is sent to the destination phone, if it subscribes to caller ID, then the number
used for routing can also be forwarded, via the signaling network to the receiver phone. The
receiver phone must have the capability to interpret the signal.
b.
Call forwarding – Allows you to have calls transferred from one phone to another where you
can be reached.
When the call request reaches the SCP and call forwarding is activated, the database is
referenced to find the forwarding number and the call is rerouted accordingly. Although the call
can be incrementally rerouted from this point, bandwidth may be conserved by rerouting the
connection again from the start to the forwarding number.
c.
Call answer – Takes voice mail message if you are on the phone or do not answer the phone.
While the phone is ringing, the number of rings must be counted somewhere in the signaling
network. After a certain number of rings (specified by the user a priori and stored in an SCP
database), if the call is not answered, it is forwarded by a mechanism similar to call forwarding to
a voice mailbox center, which handles the caller accordingly.
In another scenario, when a user is on the phone and a call arrives (assuming no call waiting),
instead of generating a busy signal, the call is forwarded automatically to the voice mailbox.
d.
Call waiting – If you are on the phone, a distinctive tone informs that you have an incoming
local or long-distance call. You can place your current call on hold, speak briefly to the
incoming party, and return to the original call.
Generally, if a line is free, a ring is sent to the dialer and receiver. If a line is busy, a busy signal
is sent back to the dialer. If call waiting is activated, a tone from the local switch is sent to the
receiving party (which briefly interrupts the current conversation).
The control plane treats this tone as a ring (that is, it sends a ring back to the dialing party and
voicemail and call forwarding may still be activated). If the user accepts the new call, the current
connection is “held” one hop upstream from the user at the local switch, and a second connection
is made from the dialing party to the user.
At this point, both calls share the line that makes up the last hop to the user. Thus, the user can
only talk to one party at a time. When one of the two connections is being used, the other is
maintained in its incomplete state while the caller is on hold.
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e.
Chapter 4 Solutions
Called-party identification – Each member of a family has a different phone number.
Incoming calls have a different ring for each phone number.
In the signaling network, multiple numbers can all be routed to the same phone. For each
number, however, a different ring signal can be sent to the phone. The phone may have only one
ringer, so to differentiate the numbers each ring can have a different duration with different length
pauses, similar to Morse code. Thus, one number may be one long ring, while the other number
may be two short rings.
53. Consider a 64 kbps connection in ISDN.
Solutions follow questions:
a.
Sketch the layers of the protocol stack in the user plane that are involved in the connection.
Assume that two switches are involved in the connection.
Only the physical layer in the user plane of the end systems and the two switches are involved in
the 64 kbps connection.
b.
Sketch the layers of the protocol stack in the control plane that are involved in the setting up
of the connection. Assume that the two switches in part (a) are involved in the call setup.
The physical, data link, network and application layers of the control plane are involved in the
setting up of the connection.
54. Explain how the control plane from ISDN can be used to set up a virtual connection in a packetswitching network. Is a separate signaling network required in this case?
Solution:
In order to setup a virtual connection in a packet switched network, the ISDN control plane must
first reserve resources during a call setup phase. The circuit does not exist as a physical entity,
but rather a sequence of entries in forwarding tables in the packet switched network. As such,
each packet in the call follows the same route through the network.
55. Identify the components in the delay that transpires from when a user makes a request for a
telephone connection to when the connection is set up. Which of these components increase as the
volume of connection requests increases?
Solution:
After the user requests a connection by dialing the phone number, the call setup procedure
involves finding a path from the source to the destination. As shown in figure 4.44a, if the two
phones are attached directly to the same switch, assuming the switch is non-blocking, the
connection can be made and the delay is equal to the processing time in the switch and the short
propagation delay of the signal.
If the two phones are not connected to the same switch, a path must be selected for the call using
the signaling network. The delay from this stage includes, processing time in each of the
signaling nodes involved, propagation delay of the signals through the network, and any routing
delay overhead. Before a connection can be made, the receiving user has to pick up the phone.
This also contributes to the overall connection setup time.
The amount of time required for the user to answer the call does not depend on the volume of
connection requests in the network. However, if there are many simultaneous requests for
connections, the signaling network will have to juggle more requests simultaneously and the
routing procedure will take longer. Although there is a fixed minimum propagation delay, as call
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Chapter 4 Solutions
requests increase, the propagation delay may also increase because the shortest path route may
not be available.
56. Discuss the fault tolerance properties of the STP interconnection structure in the signaling system
in Figure 4.52.
Solution:
The mesh structure means that any STP is connected to any other STP by more than one path.
Thus, if a fault occurs on one link, the signaling traffic can be automatically rerouted over another
path. If an STP fails, paths are available to a backup STP.
57. A set of trunks has an offered load of 10 Erlangs. How many trunks are required to obtain a
blocking probability of 2%? Use the following recursive expression for the Erlang B blocking
probability:
B (c, a ) =
aB (c − 1, a)
c + aB (c − 1, a)
B(0, a) = 1
Solution:
Given a = 10 Erlangs and PB = 2% = 0.02, 16 trunks are required to a obtain PB = 2%.
C
0
1
2
3
4
5
6
7
8
a
10
10
10
10
10
10
10
10
10
B(c,a)
1.000
0.909
0.820
0.732
0.647
0.564
0.485
0.409
0.338
c
9
10
11
12
13
14
15
16
17
B(c,a)
0.273
0.215
0.163
0.120
0.084
0.057
0.036
0.022
0.013
a
10
10
10
10
10
10
10
10
10
58. Compare the blocking probabilities of a set of trunks with offered load a = 9 and c = 10 trunks to a
system that is obtained by scaling up by a factor of 10, that is, a = 90 and c = 100. Hint: Use the
recursion in problem 57 in a spreadsheet or program.
Solution:
To find the blocking probability of trunks with offered load a = 9 and c = 10, we recursively
calculate B(10,9) starting from B(0,9). We use a similar method for the case when a = 90 and c =
100. The results of this process are summarized in the table below.
c
0
1
2
3
4
5
6
a
9
9
9
9
9
9
9
B(c,a)
1.000
0.900
0.802
0.706
0.614
0.525
0.441
c
0
1
2
3
4
5
6
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a
90
90
90
90
90
90
90
B(c,a)
1.000
0.989
0.978
0.967
0.956
0.945
0.934
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7
8
9
10
9
9
9
9
Chapter 4 Solutions
0.362
0.289
0.224
0.168
7
.
.
100
90
.
.
90
0.923
.
.
0.027
As shown, a lower probability of blocking is achieved by scaling up the system.
59. Calls arrive to a pool of 50 modems according to a Poisson process. Calls have an average
duration of 25 minutes.
Solutions follow questions:
a.
What is the probability an arriving calls finds all modems busy if the arrival rate is two calls
per minute?
For this system, the utilization a = 2/(1/25) = 50 and c = 50. The probability of blocking is:
ac
Pb =
c!
b.
aj
= 0.105
¦
j = 0 j!
c
What is the maximum arrival rate that can be handled if the maximum acceptable blocking
probability if 1%? 10%?
If the maximum acceptable blocking probability is 10% and 1%, the maximum handled arrival
rates, given a service rate of 1/25 calls per minute are:
Pb
1%
10%
Load
1.516 calls/minute
1.982 calls/minute
60. Consider dynamic nonhierarchical routing (DNHR).
Solutions follow questions:
a.
Explain how DNHR can be used to exploit the time differences between different time zones
in a continent.
In the morning in the east, the business day begins and long-distance calls are made between
cities on the East Coast. The west-coast business day has not yet begun, so the links from east
to west are relatively under-utilized.
Eastern traffic can be re-routed through the western free lines to provide additional capacity for
the traffic between eastern cities. The opposite can occur in the evening after the business day in
the east ends and the under-utilized eastern trunks are used for western traffic.
b.
Explain how DNHR can be used to exploit different business and residential activity patterns
during the day.
For local calls during business hours, traffic may be re-routed through relatively non-busy lines in
residential area to free up resources in the business areas. The opposite can be done in the
evening when most people are in the residential areas and the telephone traffic in the business
area is low volume.
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61. Suppose that setting up a call requires reserving N switch and link segments.
a.
Suppose that each segment is available with probability p. What is the probability that a call
request can be completed?
For a call to be completed, every segment in the circuit must be free. Thus, assuming that each
switch and link has a probability p of being available, and assuming that this probability is
independent of all others,
st
nd
th
N
P[every circuit free] = P[1 free]P[2 free]…P[N free] = p
b.
In allocating switch and transmission resources, explain why it makes sense to give priority to
call requests that are almost completed rather than to locally originating call requests.
For calls that are almost completed, resources are already locked up, so if the call is denied,
there is a large overhead penalty. Thus they should be given priority. To illustrate, suppose
locally originating calls are given a higher priority. One can then envision a situation where very
few completed connections exist, but all resources are used by partially completed calls.
62. Consider a cellular telephone system with the following parameters: B is the total bandwidth
available for the system for communications in both directions; b is the bandwidth required by each
channel, including guard bands; R is the re-use factor; and a is the fraction of channels used for set up.
Solutions follow questions:
a.
Find an expression for the number of channels available in each cell.
B
(1 − a )
b
Number of channels available in each cell =
channels/cell
R
b.
Evaluate the number of channels in each cell for the AMPS system.
In AMPS, there are 416 total channels, 21 of which are used for call setup. The reuse factor is 7.
Thus the number of channels per cell is:
(416 –21)/7 = 56 channels per cell
63. Consider the AMPS system in problem 62.
Solutions follow questions:
a.
How many Erlangs of traffic can be supported by the channels in a cell with a 1% blocking
probability? 5%?
The probability of blocking is given by the equation:
ac
Pb =
c!
aj
¦
j = 0 j!
c
where in the AMPS system c = 56 and a is the number of Erlangs of traffic in a cell. If the blocking
probability is 10% and 1%, the maximum number of Erlangs that can be handled is:
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Communication Networks (2nd Edition)
Load [Erlangs]
43.342
56.059
Pb
1%
10%
b.
Chapter 4 Solutions
Explain why requests for channels from handoffs should receive priority over requests for
channels from new calls. How does this change the Erlang load calculations?
In telephone conversations, service interruption is less acceptable than denial of service due to a
busy network.
The calculations in part (a) assumed equal priority for all calls. If priority is given to current calls
that are being handed off, the Erlang load calculations are more complicated. Newly attempted
calls in this scenario have a lower priority and, thus, a higher Pb.
64. Suppose that an analog cellular telephone system is converted to digital format by taking each
channel and converting it into n digital telephone channels.
Solutions follow questions:
a.
Find an expression for the number of channels that can be provided in the digital system using
the parameters introduced in problem 62.
Assuming that there will be n times more control channels in this new scheme, the number of
channels in each cell is:
B
(1 − a)
b
n
R
b.
Consider the AMPS system and assume that it is converted to digital format using n = 3.
How many Erlangs of traffic and the new system support at 1% blocking probability? 5%?
This calculation is exactly the same as that in problem 4.62, except that now the number of trunks
is tripled, so c = 160.
To support 1% blocking or 5% blocking, the traffic can be:
Pb
1%
5%
Load [Erlangs]
141.167
157.046
65. Suppose that a CDMA system has the same number of channels as the digital system in problem
64, but with a reuse factor of 1.
Solutions follow questions:
a.
How many Erlangs of traffic can be supported in each cell by this system at 1% blocking
probability? 5%?
This CDMA system has 416 channels. Assuming that 21 channels are still used for call setup
and the reuse factor is 1, each cell has 395 channels, so we have:
c = 395
a = Erlangs of traffic
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The probability of blocking in this system is given by:
Pb =
ac
c!
aj
¦
j = 0 j!
c
Thus we have:
Pb
1%
5%
b.
Load [Erlangs]
370.410
401.856
Suppose the per capita traffic generated in a city is 0.10 Erlangs during the busiest hour of the
day. The city has a population of 1 million residents and the traffic is generated uniformly
throughout the city. Estimate the number of cells required to meet the cities traffic demand
using the system in part (a).
For a probability of blocking of 1%,
Total number of cells in city
= (0.10 Erlangs/person)(1 000 000 people) / (370.410Erlangs/cell)
= 270 Cells
For a probability of blocking of 5%,
Total number of cells in city
= (0.10 Erlangs/person)(1 000 000 people) / (401.856 Erlangs/cell)
= 249 Cells
66. Consider the equipment involved in providing a call between mobile and wireline telephones.
Solutions follow questions:
a.
Sketch a diagram showing the various equipment and facilities between an originating mobile
telephone to a wireline destination telephone. Suppose first that the mobile carries analog
voice; then suppose it carries digital voice.
The mobile phone or mobile subscriber unit (MSU) transmits and receives signals to and from the
cell base station (BS) in the area where the phone is located. The cell base station exchanges the
signals with the mobile switching center (MSC) located at the mobile telephone switching office
(MTSO). All cell base stations are connected to the mobile telephone switching office through wire
or microwave channels. The mobile switching center is also connected to the local public switched
telephone network (PSTN). The mobile switching center switches the calls among the base
stations and also between the base stations and the wireline telephone network.
In analog cellular networks, the signal from the mobile phone arrives in analog form and is then
digitized for switching and transmission. In digital cellular systems the signal from the mobile user
arrives in digital form. However cellular phones use high performance speech coding formats that
require conversion PCM prior to switching in the digital telephone network.
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Mobile
Station
Base
Station
Mobile
Station
Base
Station
Chapter 4 Solutions
Mobile
Switching
Center
Mobile
Switching
Center
Base
Station
Mobile
Station
Telephone
Switch
To telephone
network
b.
Repeat part (a) in the case where the two telephone calls are mobile.
In this case once the mobile telephone switching office received the call it determines the cell
where the requested mobile phone is located and connects the call to the corresponding base
station. If the two mobile phones are in the area covered by the mobile telephone switching office,
then the signals do not traverse the public telephone network. If the two mobiles are covered by
different telephone switches then the signals must be transmitted over digital lines, and possibly
across the public telephone network.
c.
In parts (a) and (b) identify the points where a call request can be blocked during call setup
because resources are unavailable.
The call can be blocked between the mobile user and the base station if no frequency channels
are available. The call cannot likely be blocked between the Base station and the mobile
switching center since it is assumed that this link has a capacity at least equal to that of the
aggregate traffic from the base station. For this same reason, it is unlikely that the call can be
blocked between the mobile switching center and the telephone switch (in part a). The cellular
service provider leases a link to the ground-based telephone network, so it may be possible that a
call can be blocked at this point if the entire network experiences a lot of traffic.
67. Explain the signaling events that take place when call is set up and released in a cellular telephone
system. Identify the equipment and facilities involved in each phase of the call. Consider the
following cases.
Solutions follow questions:
a.
The source and destination mobile phones are in the same cell.
Whether or not the destination phone is in the same cell as the source phone, the procedure for
call setup may still be the same, because the base station does not know a priori whether the
destination phone is in its cell.
•
•
•
Mobile station sends a request in reverse setup channel along with user information such
as phone number, serial number, and authentication
Base station verifies information with MSC
MSC sends call request to all base stations
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•
•
•
•
•
b.
Chapter 4 Solutions
Base Stations broadcast request to all forward setup channels
Receiving phone replies on reverse setup channel
All base stations that receive this reply forward it to the MSC along with other information
such as the power level of the received signal. This information helps the MSC to assign
the connection channels
MSC identifies which base station is best suited to service the call for the receiving phone
The base station assigns a forward and reverse channel and the receiving phone is rung
The source and destination mobile phones are in different cells, but in the same MSC.
See answer for part (a).
c.
The source and destination mobile phones are in different cellular networks.
If the number is not stored in the MSC database, it must go through the wire-line phone network
which in turn will interface with the receiving phone’s cellular network as described above via the
MSC and the closest base station. At the cellular-wire-line network interface, the data must be
reformatted to accommodate the signaling systems of the downstream network.
68. Explain the signaling events that take place when call is handed off from one cell to another cell.
Suppose first that the two cells are under the same MSC, and then that they are under different MSCs.
Solution:
The signal level must be periodically monitored when a call is in progress. When it drops below a
certain threshold, the base station instructs the mobile station to transmit on a specific setup
channel. The base station also signals the MSC, which signals all neighboring base stations to
monitor the setup channel. The signal strength and, possibly, other information about the signal
is recorded at each base station and sent to the MSC, which uses it to decide which cell is best
suited to receive the handoff. The selected base station is informed and the old connection from
the MSC to mobile station via the original base station is broken and a new connection via the
new base station is established. The service to the mobile phone is briefly interrupted during this
process.
The process is centralized if the mobile station moves to a cell that is connected to the same
MSC as the original cell. Otherwise it is possible that a more distributed control mechanism is
used. If, after receiving the information about the signal quality at each base station, the MSC
decides that none of the cells are suitable to receive the handoff, the MSC communicates to
neighboring MSCs to locate the mobile station. If the mobile station is found in a cell of one of
these neighboring MSCs, the base station is informed and the new connection is made through
the new MSC.
69. Bob Smiley, star salesman, has just arrived in Los Angeles from his home base in Chicago. He
turns on his cell phone to contact his Los Angeles client.
a.
Sketch the sequence of events that takes place to set up his call. Assume he subscribes to
roaming service.
Bob Smiley turns on his phone, which sends user information to the MSC via the nearest base
station. The MSC contacts Bob’s home location register (in Chicago), and copies Bob’s database
information into its own visitor location register (in LA). Further, Bob’s home MSC is not aware
that his calls should be forwarded to LA. Now Bob can use his phone in the new network area.
b.
Next he calls home in Chicago to inform his wife that he forgot to take out his son’s hockey
gear from the trunk of his car and to give her the parking spot where he left the car in the
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Chapter 4 Solutions
airport (“somewhere on the third level of parking lot A”). Sketch the sequence of events that
take place to set up this call. (Don’t concern yourself with the specifics of the conversation.)
L. A.
MSC
Bob’s
Mobile
Station
Base
Station
Telephone
Network
Phone
Switch
Visitor’s
location
register
c.
Wife’s phone
In the meantime, Bob’s college roommate, Kelly, who now works in Hollywood calls Bob’s
cell phone. Note that Bob’s cell phone begins with the Chicago area code. Sketch the
sequence of events that take place to set up this call. Should this call be billed as local or long
distance?
Kelly’s long-distance service provider sees Bob’s Chicago area code and proceeds with
connecting her to Bob’s Chicago cellular network. When the call arrives at Bob’s cellular network
(in the form of signaling traffic), Bob’s home MSC checks its home location register and sees that
Bob is in LA (recall that it found out about his whereabouts in part a)). The call is then forwarded
back to LA where the connection can be made via Bob’s temporary MSC and its visitors’ location
register.
Whether Bob or Kelley should be billed long distance depends on the setup of Bob’s “roaming
service”. If the three networks involved (two cellular and one wireline) are coordinated enough,
Kelly’s final connection could be routed directly from Hollywood to Bob’s mobile station directly.
In this case it could be possible to bill them both locally.
However, it is more likely that the connection will actually be made from Kelly to Bob’s home
network and then back to L. A. By dialing long distance, Kelly implicitly agrees that she will pay
the long distance charges for a call to Chicago. By using his phone in L. A. Bob expects to pay
long distance fees for all calls he receives (even local ones) because if someone from Chicago
calls him, the call must be carried by a long distance carrier. In this case, both Bob and Kelly will
pay long distance charges. She will pay the same as if she were dialing a regular Chicago
number (which she is), and Bob will pay as if he is receiving a call from a Chicago caller, say, his
wife.
70. Compare cellular wireless networks to the local loop and to the coaxial cable television system in
terms of their suitability as an integrated access network. In particular comment on the ability to
support telephone service, high-speed Internet access, and digital television service. Consider the
following two cases:
In question 45, the local loop and coaxial cable network were compared in terms of their ability to
act as integrated access networks. Here, we examine the ability of cellular wireless networks to
provide these services.
Each type of traffic on an integrated access network has very different characteristics. Digital
television requires high bandwidth and low loss. Although the signal can be compressed using
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Communication Networks (2nd Edition)
Chapter 4 Solutions
MPEG encoding, the compressed signal still requires a minimum of 1-2 Mbps. Service can most
efficiently be provided in a broadcast network because it lacks the interactivity of phone and
Internet, and each user can be sent the same information.
Telephone service is less constrained by bandwidth. Because the vast majority of calls are
between two users, point-to-point networks are well suited for carrying phone traffic.
Current high speed Internet access has medium bandwidth requirements. IP traffic can also be
adaptive, in the sense that if need be, the connection can usually be slowed down. This affects
the quality of service, but unlike the other two (phone and TV), it does not interrupt service
completely.
a.
The band of frequencies available spans 300 MHz.
Current cellular networks have only moderate bandwidth. Cellular networks share the same
scalability problem as the coaxial television network. If the band of frequencies in the network
span 300 MHz (about 10x the current band), the bandwidth available to each user is inversely
proportional to the number of users. Of course cells can be made small, and CDMA can be used
so that the reuse factor is 1, but a bottleneck still exists. Clearly, the cellular network can support
telephone traffic, since it currently already does. Successful efforts have been made to provide
Internet service on mobile cellular devices, but due to the bandwidth restriction mentioned, the
service quality is moderate. New protocols are currently being defined that reduce bandwidth for
Internet services that run on devices of these networks.
It would be difficult to provide digital television using the above-described cellular network.
Because of the high bandwidth required, by partitioning the network into a number of parallel
channels, a broadcast network could provide digital television service.
b.
The band of frequencies available spans 2 GHz.
If the band of frequencies available spanned 2 Ghz, all of the above obstacles would be
significantly reduced. Since the aggregate bandwidth of this system is 100 times that of the
current cellular networks, most of the problems mentioned above would melt away. With a rough
calculation, assuming each cell has on the order of 100 users, and the frequency reuse factor is
1, we see that a bandwidth on the order of 10 Mbps can be provided to each user (or to each
channel in a 100-station digital television network). This is comparable to current ADSL in the
local loop network, and would likely be sufficient to serve as an integrated access network. A
difficulty with this scenario is that the region of spectrum where 2 GHz is available would be in the
28 GHz band or beyond. These frequencies tend to be suitable to point-to-point or point-tomultipoint but not to omnidirectional communications.
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Communication Networks (2nd Edition)
Chapter 5 Solutions
15. A 1 Mbyte file is to be transmitted over a 1 Mbps communication line that has a bit error rate of p = 106.
Solutions follow questions:
The file length n = 8 x 106 bits, the transmission rate R = 1 Mbps, and p = 10-6.
a.
What is the probability that the entire file is transmitted without errors? Note for n large and p very small,
(1 p)n | enp.
P[no error in the entire file] = (1 – p)n | e–np , for n >> 1, p << 1
= e-8 = 3.35 x 10-4
We conclude that it is extremely unlikely that the file will arrive error free.
b.
The file is broken up into N equal-sized blocks that are transmitted separately. What is the probability that
all the blocks arrive correctly without error? Does dividing the file into blocks help?
A subblock of length n/N is received without error with probability:
P[no error in subblock] = (1 – p)n/N
A block has no errors if all subblocks have no errors, so
P[no error in block] = P[no errors in subblock]N =((1 – p)n/N)N = (1 – p)n
So simply dividing the blocks does not help.
c.
Suppose the propagation delay is negligible, explain how Stop-and-Wait ARQ can help deliver the file in
error-free form. On the average how long does it take to deliver the file if the ARQ transmits the entire file
each time?
Refer to the following figure for the discussion.
tf
tACK
ACK/NAK
We assume the following:
x
x
x
x
x
x
t0 = basic time to send a frame and receive the ACK/NAK | ttimeout
ttotal = total transmission time until success
nf = number of bits/frame
na = number of bits per ACK
nt = number of transmissions
Pf = probability of frame transmission error
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t0 = tf + tACK = nf /R + na /R
Chapter 5 Solutions
(tprop | 0).
P[nt = i ] = P[one success after i – 1 failure] = (1 – Pf) Pf i – 1
ttotal | i transmissions = i.t0
f
E[ttotal] =
¦ it P>n
0
i 1
t
f
i @ t 0 1 Pf ¦ iPf
i 1
i 1
t 0 1 Pf / 1 Pf
2
t 0 / 1 Pf
Here, nf = n >> na thus t0 | tf = n/R ; and Pf = 1 – P[ no error] = 1 – e–np
E[total] = n/R (1 – Pf) = n/[Re–np] = 8 / (3.35 x 10–4) = 23,847 seconds = 6.62 hours!
The file gets through, but only after many retransmissions.
d.
Now consider breaking up the file into N blocks. (Neglect the overhead for the header and CRC bits.) On
the average how long does it take to deliver the file if the ARQ transmits the blocks one at a time?
Evaluate your answer for N = 80, 800, and 8000.
For 1 block Pf = 1 – Pb = 1 – (1 – p)n/N and nf = n/N
if tprop | 0 and na << n/N : t0b = nf/R = n/NR
Tb = E[ttotalb] = t0b / (1 – Pf) = n(1 – p)–n/N /NR average time to transmit one block
T = E[ttotal] = N Tb = n(1 – p)–n/N /R = 8 (1 – p) –n/N = 8 enp/N if n/N >> 1, p << 1
x
x
x
N = 80 Ÿ T | 8 e0.1 = 8.84 sec
N = 800 Ÿ T | 8 e0.01 = 8.08 sec
N = 8000 Ÿ T | 8 e0.001 = 8.008 sec
Each subblock has a higher probability of arriving without errors, and so requires fewer
retransmissions to deliver error free. The overall delay is reduced dramatically.
e.
Explain qualitatively what happens to the answer in part (d) when the overhead is taken into account.
As N increases, the effect of overhead becomes more significant because the headers constitute a
bigger fraction of each subblock.
16. Consider the state transition diagram for Stop-and-Wait ARQ in Figure 5.12. Let Pf be the probability of frame
error in going from station A to station B and let Pa be probability of ACK error in going from B to A. Suppose
that information frames are two units long, ACK frames are one unit long, and propagation and processing delays
are negligible. What is the average time that it takes to go from state (0,0) to state (0,1)? What is the average time
that it then takes to go from state (0,1) to state (1,1)? What is the throughput of the system in information
frames/second?
Solution:
We know that Pf is the probability of frame error and Pa is the probability of ACK error. We assume
that:
x
X is the random variable that represents the number of trials before a successful transmission of
a frame. Each unsuccessful trial requires a timeout for retransmission. We assume that the
timeout time is set to be equal to frame transmission time plus ACK transmission time.
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Communication Networks (2nd Edition)
x
x
Chapter 5 Solutions
Y is the random variable that represents the number of trials before a successful transmission of
an ACK. An ACK error will require a new successful retransmission of the frame for next ACK
transmission. An ACK is not sent until a new retransmitted frame arrives at the receiver.
X and Y follow a geometric random-variable distribution.
T1 = Average time to go from (0,0) to (0,1) = (Tf + Ta) E(X) + Tf
T 1 (2 1)
Pf
P
ª P
º
2 2« f 1» f
1 Pf
¬1 Pf ¼ 1 Pf
P
2
f
1 Pf 1 Pf
2 Pf
1 Pf
T2 = Average time to go from (0,1) to (1,1) = T1E(Y) +Ta
ª Pa º
Ta
T 2 T1 «
¬1 Pa »¼
ª 2 Pf º ª Pa º
«
»«
» 1
«¬ 1 Pf »¼ ¬1 Pa ¼
Throughput = Frame Time / Expected Total Transmission Time = 2 / (T1 + T2)
Pf
Pa
T1
T2
Throughput
0.2
0.1
2.75
1.3055
0.4931
0.02
0.01
2.06
1.0208
0.6492
0.002
0
0.001
0
2.006
2.0000
1.0020
1.0000
0.6649
0.6667
18. A 64-kilobyte message is to be transmitted from the source to the destination. The network limits packets to a
maximum size of two kilobytes, and each packet has a 32-byte header. The transmission lines in the network have a
bit error rate of 106, and Stop-and-Wait ARQ is used in each transmission line. How long does it take on the
average to get the message from the source to the destination? Assume that the signal propagates at a speed of 2 x
105 km/second.
1000 km
1000 km
Switch
Source
1.5 Mbps
Destination
1.5 Mbps
Solution:
Message Size
65536 bytes
Max Packet Size
2048 bytes
Packet Header
32 bytes
Available for info
2016 bytes
# of packets needed 32.51 packets
Total
33 packets
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Chapter 5 Solutions
bit error rate
1E-06
bits/packet
16384
Probability of error in packet 0.016251
Propagation speed
2E+05 Km/s
Distance
1000 Km
Bandwidth
1.5
Mb/s
1 – (1– bit_error_rate) ^ (bits/packet)
We assume that the ACK error, the ACK time, and processing time are negligible.
Tprop =
Tf =
T0 =
Pf =
distance / propagation speed = 0.0050 s
packet size / bandwidth = 0.0109 s
Tprop + Tf = 0.0159 s
probability of error in packet = 0.016251
E[Ttotal]=
T0/ (1 - Pf) = 0.0162
There is pipelining effect that occurs as follows: After the first packet arrives at switch 1, two
transmissions take place in parallel. The first packet undergoes stop-and-wait on the second link
while the second packet undergoes stop-and-wait in the first link. The packet arriving at the switch
cannot begin transmission on the next link until the previous packet has been delivered, so there is an
interaction between the transmission times of the two packets. We will neglect this effect. The time
to send every packet over two links is then the initial packet transmission time + 33 additional packet
timess, and so the average time is E[Ttotal] * 34 = 0.522 seconds.
19. Suppose that a Stop-and-Wait ARQ system has a time-out value that is less than the time required to receive an
acknowledgment. Sketch the sequence of frame exchanges that transpire between two stations when station A sends
five frames to station B and no errors occur during transmission.
Solution:
Timeout
Fr 0
Fr 0
ACK
1
Fr 1
Fr 1
ACK
1
ACK
0
Fr 0
Fr 0
ACK
0
ACK
1
Fr 1
Fr 1
ACK
1
ACK
0
Fr 0
ACK
0
ACK
1
20. The Trivial File Transfer Protocol (RFC 1350) is an application layer protocol that uses the Stop-and-Wait
protocol. To transfer a file from a server to a client, the server breaks the file into blocks of 512 bytes and sends
these blocks to the client using Stop-and-Wait ARQ. Find the efficiency in transmitting a 1 MB file over a 10 Mbps
Ethernet LAN that has a diameter of 300 meters. Assume the transmissions are error free and that each packet has
60 bytes of header attached.
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Communication Networks (2nd Edition)
Chapter 5 Solutions
Solution:
The propagation delay in an Ethernet LAN is negligible compared to the total transmission time of a
packet from start to finish. Ignoring processing time and using the terminology in the chapter, we
have:
to = tf + tack =
8(512 60)
64
6
10 u 10
10 u 10 6
n f no
Ko =
R
0
eff
R
8 u 512
4.64 u 10 4
10 u 10 6
to
R
4.64 u 10 4
0.8828
88.3%
One more source of overhead occurs because the last packet is not full. However, this additional
overhead accounts for a very small fraction of the total overhead and does not affect the above result.
21. Compare the operation of Stop-and-Wait ARQ with bidirectional Go-Back-N ARQ with a window size of 1.
Sketch out a sequence of frame exchanges using each of these protocols and observe how the protocols react to the
loss of an information frame and to the loss of an acknowledgment frame.
Solution:
The figure below shows that the bidirectional Go-Back-N ARQ recovers from errors in the same time
that Stop-and-Wait:
Stop & Wait
timeout
Fr0
Fr1
timeout
Fr1
ACK1
ACK0
Go Back N
ACK1
Fr1
Fr0
Fr0
ACK1
ACK1
timeout
ACK0
Fr0
Fr1
ACK0
Fr0
ACK1
timeout
ACK1
Fr1
Fr0
Fr0
ACK1
timeout
ACK0
Fr0
Fr0
ACK1
Fr0
ACK1
timeout
22. Consider the various combinations of communication channels with bit rates of 1 Mbps, 10 Mbps, 100 Mbps,
and 1 Gbps over links that have roundtrip times of 10 msec, 1 msec, and 100 msec.
Solutions follow questions:
a.
Find the delay-bandwidth product for each of the 12 combinations of speed and distance.
Delay-bandwidth (Megabits)
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Bit Rate
Mbps
1
10
100
1000
b.
Chapter 5 Solutions
Round Trip Time (msec)
10
1
0.1
0.01
0.001
1.0
0.10
0.010
10.0
1.00
0.100
100.0
10.00
1.000
100
Suppose 32-bit sequence numbers are used to transmit blocks of 1000 bytes over the above channels. How
long does it take for the sequence numbers to wrap around, that is, to go from 0 up to 2m?
Block 1000 bytes
Sequence
32 bits 4294967296
Time for the sequence number to wrap around (sec)
Frame Time * 232=4294967296*8*1000/R seconds
Bit Rate
Mbps
1
10
100
1000
c.
Round Trip Time (msec)
100
10
1
34359738.37 34359738.37 34359738.37
3435973.84 3435973.84 3435973.84
343597.38
343597.38
343597.38
34359.74
34359.74
34359.74
Now suppose the 32-bit sequence numbers are used to count individual transmitted bytes. How long does it
take for the sequence numbers to wrap around?
Time for the sequence number to wrap around (sec)
Byte Time * 232=4294967296*8/R seconds
Bit Rate
Mbps
1
10
100
1000
Round Trip Time (msec)
100
10
1
34359.74
34359.74
34359.74
3435.97
3435.97
3435.97
343.60
343.60
343.60
34.36
34.36
34.36
The sequence number wraps around in much shorter time. At 1 Gbps the sequence number wraps
around in only 34.36 seconds.
23. Consider a bidirectional link that uses Go-Back-N with N = 7. Suppose that all frames are one unit long and
that they use a time-out value of 2. Assume the propagation is 0.5 unit and the processing time is negligible.
Assume the ACK timer is one unit long. Assuming stations A and B begin with their sequence numbers set to zero,
show the pattern of transmissions and associated state transitions for the following sequences of events:
Solutions follow questions:
Go-Back-N with N = 7, ttimeout = 2, tprop = 0.5, tACK = 1, tf = 1. Each tick represents one half of a unit
of time. The ACK timer delays the sending of an ACK to provide an opportunity for piggybacking.
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a.
Chapter 5 Solutions
Station A sends six frames in a row, starting at t = 0. All frames are received correctly.
b.
{5}
Ack5
Fr7
Ack0
Rn=0
AckTo
Rn=7
Rn=6
Rn=5
Rn=6
Rn=5
Rn=5
AckTo
Ack6
Fr6
AckTo
Rn=4
Rn=3
Fr5
{7}
{4,5,6}
{4,5}
Fr5
Ack4
Fr4
AckTo
Rn=2
Rn=1
Rn=0
AckTo
Rn=4
Rn=3
Rn=2
{2,3}
Fr3
Ack2
Fr2
Fr3
AckTo
{2,3,4}
Rn=2
Rn=1
AckTo
{0,1,2}
Rn=0
{0}
{0,1}
Fr1
Fr4
Fr3
Ack5
Fr2
Ack3
Fr0
Ack2
Fr1
AckTo
Fr0
{3,4}
To3
To0
Fr0
{3,4}
{2,3}
{2}
{0,1}
{0}
{0,1}
The first figure below shows that the timeout value is too short and cause needless retransmissions.
The second figure below shows that by increasing the timeout value to 3 units, transmissions occur
smoothly and efficiently.
Station A sends six frame in a row, starting at t = 0. All frames are received correctly, except frame 3 is
lost.
The following two figures show the sequence of frame exchanges with timeout values of 2 and 3
units. The system with the shorter timeout value initiates the recovery from error sooner, but cannot
advance the window fast enough due to the short timeout value.
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{5}
{5}
Fr5
Ack6
Rn=6
Ack5
Fr5
Rn=5
AckTo
Rn=4
Rn=3
AckTo
{5}
{3,4,5}
{3,4}
Fr4
Ack3
Fr3
Ack3
Rn=3
Rn=5
Rn=4
{3,4}
Fr4
X
Rn=3
{3,4}
Rn=3
Rn=3
{2,3,4}
{2,3}
Fr3
AckTo
Rn=2
Rn=1
Fr3
AckTo
Ack2
Fr2
{3,4}
{3,4}
Fr4
Ack5
Ack3
X
Rn=3
Rn=2
{0,1,2}
Rn=2
Fr1
AckTo
Rn=0
Fr3
AckTo
{0,1}
Rn=1
{0}
Rn=0
Fr0
Fr4
Fr3
Rn=3
Fr2
To3
Ack3
Fr0
Ack2
Fr1
AckTo
c.
{3,4}
To3
To0
Fr0
{3,4}
{2,3}
Chapter 5 Solutions
{2}
{0,1}
{0}
{0,1}
Communication Networks (2nd Edition)
Station A sends six frames in a row, starting at t = 0. Station B sends six frames in a row starting at t =
0.25. All frames are received correctly.
This problem shows that when piggybacking is used the timeout has too be somewhat longer. In the
following figure we see that timeouts occur repeatedly causing needless retransmission of frames that
have already arrived at the receiver. A timeout value of 4 is required to assure a smooth sliding
forward of the transmitter’s send window.
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3,4
4,4
{3} Rn=4
{3,4} Rn=4
{4} Rn=4
5,4
{5} Rn=6
2,3
4,5
5,6
4,5
5,6
{5} Rn=6
3,2
{4,5} Rn=6
2,2
5,4
{4,5} Rn=5
{4,5} Rn=6
1,2
4,4
{4,5} Rn=5
0,1
3,4
{4,5} Rn=4
{4,5} Rn=5
2,3
{2,3} Rn=4
{4} Rn=4
{4,5} Rn=4
3,2
{3} Rn=4
{3,4} Rn=4
2,2
{2,3} Rn=2
{2,3} Rn=3
{2,3} Rn=4
{2} Rn=2
{2,3} Rn=2
1,2
{1,2} Rn=2
{2} Rn=2
{2,3} Rn=3
{1} Rn=2
{1,2} Rn=2
1,0
{0,1} Rn=0
{0,1} Rn=1
{0}, Rn=0
0,0
0,1
{1} Rn=2
1,0
Chapter 5 Solutions
{0,1} Rn=2
0,0
{0,1} Rn=2
{0}, Rn=0
{0,1} Rn=0
{0,1} Rn=1
Communication Networks (2nd Edition)
25. Consider the Go-Back-N ARQ protocol.
Solutions follow questions:
a.
What can go wrong if the ACK timer is not used?
When no traffic arrives at a receiver during bidirectional Go-Back-N ARQ, and the receiver has to
send an ACK, it usually sends the ACK after the ACK timer expires. If the ACK timer is not used,
there are only two options remaining:
1) The ACK must be sent immediately (that is, use piggybacking only if frame already available)
Although this will function correctly, it is an inefficient use of bandwidth in the general case.
2) The ACK must only be sent if it can be piggybacked
This is problematic if traffic arrives sporadically. The sender will wait a long time until a piggyback
opportunity arises.
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b.
Chapter 5 Solutions
Show how the frame timers can be maintained as an ordered list where the time-out instant of each frame is
stated relative to the time-out value of the previous frame.
Assume that the timer counts down from ttimeout. In order to have a separate timer for each frame, we
need not implement N timers. Only the oldest frame can timeout. The system can save, for each
frame, an arrival offset time that is related to the frame that preceded it and place these offsets in an
ordered list based on the frame sequence numbers. If an ACK for the oldest frame arrives, the
system simply increments the timer by the offset of the following frame in the list. If an ACK for any
other frame arrives, the timer is incremented by the sum of all of the offsets in the list that are up to
this newly acknowledged frame.
c.
What changes if each frame is acknowledged individually instead of by using a cumulative
acknowledgment (Rnext acknowledges all frames up to Rnext –1)?
If each frame needs to be acknowledged individually, then the number of ACK messages will
increase and the rate at which the transmission window can be increased will be reduced.
29. Three possible strategies for sending ACK frames in a Go-Back-N setting are as follows: send an ACK frame
immediately after each frame is received, send an ACK frame after every other frame is received, and send an ACK
frame when the next piggyback opportunity arises. Which of these strategies are appropriate for the following
situations?
Solutions follow questions:
a.
An interactive application produces a packet to send each keystroke from the client; the server echoes each
keystroke that it receives from the client.
Since each keystroke is echoed, there will always be a piggyback opportunity. Thus, the piggyback
method should be used. Indeed, the echo packet constitutes an acknowledgment.
b.
A bulk data transfer application where a server sends a large file that is segmented in a number of full-size
packets that are to be transferred to the client.
The upstream traffic to a server is generally much less than the downstream traffic. Thus, the
piggybacking method is non-ideal in this case. If the channel has a low probability of error, the
alternating ACK method is better, as it saves bandwidth. However, if the connection causes frequent
errors, every frame should be acknowledged. Additional overhead traffic is caused by the ACK
frames, but will compensated by the bandwidth savings that will arise when the errors are discovered
more quickly.
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Chapter 5 Solutions
51. Explain why framing information is required even in the case where frames are of constant length.
Solution:
To detect frame boundaries based on the frame length, the first frame must be detected correctly,
which requires synchronization information. Also the receiver must have exact timing of the
transmitter which is not practical and as a result the bit synchronization may be lost. The
synchronization may be lost because of bit errors or loss of data, or loss of signal in the link.
52. Perform the bit stuffing procedure for the following binary sequence: 1101111111011111110101.
Solution:
The inserted stuff bits are underlined.
1101111111011111110101 Æ 110111110110111110110101
53. Perform bit de-stuffing for the following sequence: 11101111101111100111110.
Solution:
The removed stuff bits are indicated by a ‘-‘.
11101111101111100111110 Æ 111011111-11111-01111154. Consider the PPP byte stuffing method. What are the contents of the following received sequence of bytes after
byte destuffing:
0x7D 0x5E 0xFE 0x24 0x7D 0x5D 0x7D 0x5D 0x62 0x7D 0x5E
Solution:
0x7D 0x5E 0xFE 0x24 0x7D 0x5D 0x7D 0x5D 0x62 0x7D 0x5E
o 0x7E 0xFE 0x24 0x7D 0x7D 0x62 0x7E
63. Use the Ethereal packet capture tool to analyze the exchange of packets during PPP setup using ad dial up
connection to a local ISP. Start the Ethereal packet capture and then initiate the connection to the ISP. The
number of captured packets will stop increasing after some point.
Solutions follow questions:
a.
Analyze the options fields of the packets during the LCP negotiations. You may need to consult RFC 1662
to interpret some of the packet contents.
b.
Examine the packet contents during the PAP exchange. Repeat the packet capture using CHAP.
The screen capture below shows a sequence of LCP and NCP negotiations for PPP. The twp
stations exchange LCP messages where they propose and reject various configuration options. The
highlighted frame (9) shows the final ACK accepting a configuration. It can be seen from the middle
pane that PAP authentication will be used. Frame 10 sends the password request, and frame 11
sends the password.
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Communication Networks (2nd Edition)
c.
Chapter 5 Solutions
Analyze the IPCP packet exchange. What IP addresses are configured during the exchange? What other
options are negotiated? You may wish to consult RFC 1332 for this part.
The screen capture below shows a sequence the final IPCP ACK message confirming the use of
protocol field compression and various IP addresses.
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64. Explain the differences between PPP and HDLC.
Solution:
HDLC can support various data transfer modes, supports multipoint links and point to point links, and
can implement error control and flow control mechanisms. PPP uses HDLC-like frames but does not
use error control and flow control protocols. Instead PPP supports powerful link and network control
protocols.
PPP is character based and can be implemented on any physical layer, HDLC is bit based and can
be implemented only on bit synchronous physical layer.
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Solutions to Chapter 6
1. Why do LANs tend to use broadcast networks? Why not use networks consisting of multiplexers and switches?
Solution:
The computers in a LAN are separated by a short distance (typically < 100m) so high speed and
reliable communication is possible using a shared broadcast medium. The cost of the medium is
negligible and the overall cost is dominated by the cost of the network interface cards in each
computer. In addition, the LAN users usually belong to the same group where all users are generally
trusted, so broadcast does not pose much security danger.
The original reason for avoiding a multiplexer and switch approach to LANs is that a centralized,
expensive “box” is required. The availability of Application Specific Integrated Circuits (ASICs) has
reduced the cost of switching boxes and made switch-based LANs feasible, and in some
environments the dominant approach.
2. Explain the typical characteristics of a LAN in terms of network type, bit rate, geographic extent, delaybandwidth product, addressing, and cost. For each characteristic, can you find a LAN that deviates from the typical?
Which of the above characteristics is most basic to a LAN?
Solution:
x
x
x
x
x
x
x
Type: broadcast network in a bus, ring or star topology.
Bit rate: from 1Mbps to 100 Mbps.
Delay-bandwidth product: small.
Addressing: flat.
Geographical extent: up to 1000 m (small).
Cost: low.
10 Gbps Ethernet deviates from the above characteristics in that it is non-broadcast, of much
higher bit rate, large delay-bandwidth product, larger geographic extent and high cost.
The most basic characteristic of a LAN is small geographical extent.
3. Compare the two-channel approach (Figure 6.4) with the single-channel approach (Figure 6.5) in terms of the
types of MAC protocols they can support.
Solution:
Figure 6.4 consists of two unidirectional channels, one outbound from a central node to secondary
nodes, and another inbound from the secondaries to the central node. The bandwidth that is
available in each direction is fixed. This arrangement can support polling protocols as well as
contention protocols in the inbound direction. Figure 6.5 provides a single channel that is shared by
all stations. This arrangement also supports polling and contention-type MAC protocols. The
bandwidth available in each direction can be controlled dynamically.
4. Suppose that the ALOHA protocol is used to share a 56 kbps satellite channel. Suppose that frames are 1000 bits
long. Find the maximum throughput of the system in frames/second.
Solution:
Maximum throughput for ALOHA = 0.184
Maximum throughput in frames/sec = (56000 bits/sec) x (1 frame/1000 bits) x 0.184 = 10.304
The maximum throughput is approximately 10 frames/sec.
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9. In a LAN, which MAC protocol has a higher efficiency: ALOHA or CSMA-CD? What about in a WAN?
Solution:
The maximum efficiency achieved by the Slotted ALOHA is 0.368. The efficiency of CSMA-CD is
given by 1/(1 + 6.4a), and is sensitive to a = tpropR/L, the ratio between delay-bandwidth product and
frame length.
In a LAN environment, the end-to-end distance is around 100m and the transmission rates are
typically 10Mbps, 100Mbps and 1Gbps (See Table 6.1). An Ethernet frame has a maximum length of
1500 bytes = 12,000 bits.
The table shows the efficiency of CSMA-CD at various transmission rate. Assume L = 12,000 bits and
propagation speed of 3 x 108.
10 Mbps
100 Mbps
1 Gbps
a
3 x 10-4
3 x 10-3
3 x 10-2
Efficiency
0.998
0.981
0.839
Note however that if shorter frame sizes predominate, e.g. 64 byte frames, then a increases by a
factor of about 20. According to the above formula the efficiency of CSMA-CD at 1 Gbps then drops
to about 0.7. The situation however is worse in that the minimum frame size at 1 Gbps needs to be
extended to 512 bytes, as discussed in page 436 of the text.
In a WAN environment d is larger. Assuming 100 Km, a is larger by a factor of 103 resulting in an
efficiency of 0.36, 0.05, and 0.005 respectively for 10 Mbps, 100 Mbps, and 1 Gbps transmission
rates. In the case of 10 Mbps transmission rate the efficiency of CSMA-CD is close to the efficiency of
ALOHA but in the other two cases it is much less than ALOHA.
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36. Suppose that 80 percent of the traffic generated in the LAN is for stations in the LAN, and 20 percent is for
stations outside the LAN. Is an Ethernet Hub preferable to an Ethernet switch? Does the answer change if the
percentages are reversed?
Solution:
The difference between a hub and a switch is that in the hub frame are broadcast to all lines, while in
a switch, frames are forwarded to another collision domain only if the destination is in that domain.
When 80% of the traffic is local, the switch will only forward 20% of frames to other collision domains,
thus increasing the bandwidth available on those domains. If 80% of the traffic is to other collision
domains, then the switch forward more traffic to other domains. If there is only one other domain,
then the switch forwards almost as much traffic as a hub would and is thus ineffective in containing
broadcast traffic. On the other hand, if the switch has multiple broadcast domains, then the amount
of traffic forwarded from the switch will be less than that forwarded by a hub.
37. Calculate the parameter a and the maximum throughput for a Gigabit Ethernet hub with stations at a 100-meter
distance and average frame size of 512 bytes; 1500 bytes; and 64,000 bytes.
d = 100
Tprop = 0.0000005
R = 1.00E+09
a
Throughput
512
0.12207
0.56
1500
0.041667
0.788525
64000
0.000977
0.993754
42. Use IEEE 802.3 and IEEE 802.11 to discuss three differences between wired and wireless LANs.
Solution:
Error rate: Unlike wired LANs, wireless LANs have high error rate due to interference and noise.
Wireless LANs need to implement ARQ and/or error correction to increase the reliability of the
communication channel.
Station mobility: Unlike wired LANs where stations connected to the LANs are static, in wireless
LANs, the stations can be mobile and portable. Wireless LAN protocols may have to implement
dynamic traffic routing and service handoff when the station moves from one service area to another.
Collision detection: Collision detection is not effective in wireless LANs due to the hidden station
problem. Consequently, the sender must wait for explicit acknowledgment (e.g. RTS/CTS) from the
receiver to know whether or not a frame has been received. The wireless LAN protocol implements a
collision avoidance algorithm rather than the collision detection in wired LAN, and the delay in the
contention period is longer than thee round-trip delay of 2tprop of wired LAN because of waiting for the
receiver’s acknowledgment.
Other differences:
Security: In a wired LAN, the transmission medium is usually physically secure. In a wireless LAN,
any device within the geographic transmission area can intercept the transmissions. To provide data
security, wireless LANs need to implement encryption at the expense of higher cost and reduced
performance.
Power consumption: Portable and mobile devices are usually battery powered, and thus have limited
power capacity. The wireless LAN protocol must be designed to be power efficient.
All these issues are addressed in the IEEE 802.11 wireless LAN protocol.
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43. For data packet radio networks, discuss the advantages and disadvantages of providing reliability by (a)
implementing error correction at the physical layer, (b) implementing error control as part of the MAC layer, and (c)
implementing error control at the LLC layer.
Solution:
Error correction at the physical layer uses up bandwidth in the form of check bits and adds complexity
in terms of hardware for detecting and correcting errors. If the bit error rate is low, ARQ at a higher
layer may be simpler and more efficient. However, if the bit error rate is high, error correction may
become essential to be able to communicate at all over the radio medium.
Error control at the MAC layer is the preferred approach if the error rate is not too high. The number
of check bits required for error detection is less than for error correction. Bandwidth is “wasted” only
on retransmissions, not in a large number of check bits in every transmission. The implementation of
ARQ is much simpler than complex error correction.
Error control at the LLC layer provides flexibility when operating over a MAC sublayer. The MAC
sublayer can provide connectionless service and the LLC can add reliability for those network layers
that require it.
44. Consider the distributed coordination function in IEEE 802.11. Suppose that all packet transmissions are
preceded by a RTS-CTS handshake. Find the capacity of this protocol following the analysis used for CSMA-CD.
Solution:
In IEEE 802.11 DCF with RTS-CTS handshake, the sender stations contend for the channel by
sending a RTS frame to the receiver, and it successfully captures the channel only when it receives a
CTS frame from the receiver.
The sender does not know if it has succeeded until time = DIFS + XRTS + SIFS + XCTS. By this time, if
no CTS frame arrives, the sender knows it has failed and will execute a backoff. This duration is
similar to 2tprop in CSMA-CD. Therefore, in CSMA-CA, the time can be divided in contention slots of
size (DIFS + XRTS + SIFS + XCTS).
Src
DIFS
RTS
SIFS
SIFS CTS
Dest
Contention slot
DIFS
Data
SIFS
ACK
Data transmission
RTS
SIFS CTS
Contention slot
Similar to CSMA-CD, suppose all stations are contending for the channel and have a probability of p
to transmit RTS during a contention slot. Then the maximum probability of success is 1/e as the
number of station increases, and the average # of contention slots in a contention period is e.
The maximum throughput occurs when all of the channel time is spent in transmission period followed
by contention intervals.
ρ max =
X data
X data + 2 SIFS + X ack + e( DIFS + X rts + SIFS + X cts )
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47. Why is error control (ARQ and retransmission) included in the MAC layer in IEEE 802.11 and not in IEEE
802.3?
Solution:
Error control is required in the MAC in IEEE 802.11 because of the noise and interference in the
wireless medium. In contrast, the wired medium for IEEE 802.3 protocol has very low error rates.
52. Six stations (S1-S6) are connected to an extended LAN through transparent bridges (B1 and B2), as shown in
the figure below. Initially, the forwarding tables are empty. Suppose the following stations transmit frames: S2
transmits to S1, S5 transmits to S4, S3 transmits to S5, S1 transmits to S2, and S6 transmits to S5. Fill in the
forwarding tables with appropriate entries after the frames have been completely transmitted.
S1
S2
S3
LAN1
S4
S6
S5
LAN2
port 1
B1
Station
port 2
Port
LAN3
port 1
B2
Station
port 2
Port
Solution:
Station
S2
S5
S3
S1
Port
1
2
2
1
Station
S2
S5
S3
S1
S6
Port
1
2
1
1
2
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Communication Networks (2nd Edition)
Chapter 7 Solutions
Solutions to Chapter 7
7.1. Explain how a network that operates internally with virtual circuits can provide connectionless service.
Comment on the delay performance of the service. Can you identify inefficiencies in this approach?
Solution:
The connection-oriented (co) network can present a network sublayer interface to the upper layer to
make itself appear as a connectionless (cl) network. The upper layer sends packets in connectionless
fashion, but the network resolves the packet destination’s cl address into the corresponding co
address (e.g. ATM address), and then establishes a virtual circuit between the source and
destination. Any subsequent packets with the same source and destination are transmitted through
this virtual circuit. The interface hides the internal operation of the network from the upper layer.
The cl-to-co address resolution is performed for the transmission of every packet, and hence incurs
extra processing delay. Address caching can speed up the process, but cannot eliminate the delay.
The first cl packet incurs extra delay because of the time required to set up the virtual circuit.
This cl-service-over-co-network approach is inefficient in that the QoS support of the connectionoriented network cannot be made available to the network layer.
ATM LAN emulation provides an example that involves providing connectionless service over a
connection-oriented service.
7.2. Is it possible for a network to offer best-effort connection-oriented service? What features would such
a service have, and how does it compare to best-effort connectionless service?
Solution:
Best-effort connection-oriented service would involve the transfer of packets along a pre-established
path in a manner that does not provide mechanisms for dealing with the loss, corruption or misdelivery of packets. Best-effort connection-oriented service would require some means for establishing
a path prior to the transfer of packets. Best-effort connectionless service would involve the transfer of
packets in a datagram fashion, where routing decisions are made independently for each packet.
The path setup requirement makes connection-oriented service more complex than connectionless
service. On the other hand, once a path is established, less processing is required to decide how a
packet is to be forwarded. Connectionless service is more robust than connection-oriented service
since connectionless service readily reroutes packets around a failure while VC service requires that
new paths be established.
7.3. Suppose a service provider uses connectionless operation to run its network internally. Explain how the
provider can offer customers reliable connection-oriented network service.
Solution:
To provide connection-oriented network service, an upper sublayer in the network layer at the edge of
the network can establish logical connections across the connectionless network by setting up state
information (for example, packet sequence number). The logical connection must be set up before
packets can be transported, and each packet is assigned a sequence number. Using the sequence
number, the upper sublayer entities can acknowledge received packets, determine and retransmit lost
packets, delete duplicate packets, and rearrange out-of-order packets, hence providing reliable
connection-oriented network service.
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7.4. Where is complexity concentrated in a connection-oriented network? Where is it concentrated in a
connectionless network?
Solution:
The complexity in connection-oriented networks revolves around the need to establish and maintain
connections. Each node must implement the signaling required by the connection establishment
process; each node must also maintain the state of the node in terms of connections already
established and transmission resources available to accommodate new connections. End systems
must be capable of exchanging signaling information with the network nodes to initiate and tear down
connections. A connection oriented network must also include routing to select the paths for new
connections.
Connectionless networks only require that nodes forward packets according to its routing tables. End
systems only need to place network address information in the packet headers. The complexity of
connectionless networks revolves around routing tables. Routing tables may be static and set up by
a network administrator, or they may be dynamic and involve processing and exchange of link state
information among nodes.
7.18. An audiovisual real-time application uses packet switching to transmit 32 kilobit/second speech and
64 kilobit/second video over the following network connection.
1 km
Work
Station
3000 km
Switch 1
10 Mbps
1000 km
Switch 2
45 Mbps
or 1.5 Mbps
1 km
Work
Station
Switch 3
45 Mbps
or 1.5 Mbps
10 Mbps
Two choices of packet length are being considered: In option 1 a packet contains 10 milliseconds of
speech and audio information; in option 2 a packet contains 100 milliseconds of speech and audio
information. Each packet has a 40-byte header.
Solutions follow questions:
(a)
For each option find out what percentage of each packet is header overhead.
Option 1:
10 ms of speech gives 32 x 10 bits of information
10 ms of audio gives 64 x 10 bits of information
40-byte header produces 40 x 8 bits of header
Therefore, the packet size is 1280 bits, with an overhead of 25%.
Option 2:
100 ms of speech gives 32 x 102 bits of information
100 ms of audio gives 64 x 102 bits of information
40-byte header produces 40 x 8 bits of header
Therefore, the packet size is 9920 bits, with an overhead of 3.225%.
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(b) Draw a time diagram and identify all the components of the end-to-end delay. Keep in mind that a
packet cannot be sent until it has been filled and that a packet cannot be relayed until it is
completely received (that is, store and forward). Assume that bit errors are negligible.
The end-to-end delay contains the following components: 1. a packetization delay Ps consisting of
the time it take a source to generate enough information to fill a packet; the first byte in the packet is
delayed by this amount: 2. queueing delay, frame transmission delay, and propagation delay at each
hop.
We assume that a message is the same as a packet and that
delay = Ps + P1 + P2 + P3 + P4 + T1 + T2 + T3 + T4
Propagation
delays
Transmission through
four hops
7.29. Consider the datagram packet network in Figure 7.25. Reconstruct the routing tables (using
minimum hop routing) that result after node 4 fails. Repeat if node 3 fails instead.
Solution:
1
2
3
4
5
6
3
2
3
2
3
6
1
2
4
5
6
1
1
6
6
1
2
3
4
5
3
5
3
5
2
1
3
4
5
6
5
1
1
5
5
1
2
3
4
6
2
2
6
6
If node 3 fails:
6
1
2
3
4
5
6
2
4
2
4
4
1
2
3
5
6
1
2
5
5
1
2
3
4
5
2
1
3
4
5
6
5
5
5
5
5
1
4
5
5
1
2
3
4
6
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7.30. Consider the following six-node network. Assume all links have the same bit rate R.
B
A
E
D
C
F
Solutions follow questions:
(a)
Suppose the network uses datagram routing. Find the routing table for each node, using minimum
hop routing.
Node A:
destination
B
C
D
E
F
next node
B
C
B
B
B
Node B:
destination
A
C
D
E
F
next node
A
D
D
D
D
Node C:
destination
A
B
D
E
F
next node
A
A
D
D
D
Node D:
destination
A
B
C
E
F
next node
B
B
C
E
F
Node E:
destination
A
B
C
D
F
next node
D
D
D
D
D
Node F:
destination
A
B
C
D
E
next node
D
D
D
D
D
(b) Explain why the routing tables in part (a) lead to inefficient use of network bandwidth.
Minimum-hop based routing causes all traffic flow through the minimum hop path, while the
bandwidth of other paths of equal and longer distance are unused. It is inefficient in its usage of
network bandwidth and likely to cause congestion in the minimum-hop paths.
For example, in the routing table of node A, the traffic flows from A to E and A to F both pass through
node B, and leave the bandwidth of link AC and CD unused.
(c)
Can VC routing give better efficiency in the use of network bandwidth? Explain why or why not.
Yes. During the VC setup, the links along the path can be examined for the amount of available
bandwidth, if it below a threshold, the connection will be routed over alternate path. VC routing allows
more even distribution of traffic flows and improves the network bandwidth efficiency.
(d) Suggest an approach in which the routing tables in datagram network are modified to give better
efficiency. Give the modified routing tables.
Assign a cost to each link that is proportional to the traffic loading (number of connections) of the link.
A higher cost is assigned to more congested links. Thus the resulting routing table avoids congested
links and distributes traffic more evenly.
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The changes in the routing table are in bold and italicized.
Node A:
destination
B
C
D
E
F
next node
B
C
B
C
C
Node B:
destination
A
C
D
E
F
next node
A
A
D
D
D
Node C:
destination
A
B
D
E
F
next node
A
A
D
D
D
Node D:
Destination
A
B
C
E
F
next node
B
B
C
E
F
Node E:
destination
A
B
C
D
F
next node
D
D
D
D
D
Node F:
destination
A
B
C
D
E
next node
D
D
D
D
D
7.32. Consider the network in Figure 7.30.
Solutions follow questions:
(a)
Use the Bellman-Ford algorithm to find the set of shortest paths from all nodes to destination node
2.
Iteration
Initial
1
Node 1
( -1 , f )
(2,3)
(3,f)
(4,f)
Node 3
( -1 , f )
(1,f)
(4,f)
(6,f)
2
(2,3)
(3,f)
(4,6)
(1,5)
(4,3)
(6,f)
3
(2,3)
(3,5)
(4,6)
(1,5)
(4,3)
(6,7)
4
(2,3)
(3,5)
(4,6)
(1,5)
(4,3)
(6,5)
Node 4
( -1 , f )
(1,f)
(2,1)
(3,f)
(5,f)
(1,8)
(2,1)
(3,f)
(5,7)
(1,8)
(2,1)
(3,5)
(5,7)
(1,8)
(2,1)
(3,5)
(5,7)
Node 5
( -1 , f )
(2,4)
(4,f)
(6,f)
Node 6
( -1 , f )
(3,f)
(5,f)
(2,4)
(4,4)
(6,f)
(3,f)
(5,6)
(2,4)
(4,4)
(6,8)
(3,4)
(5,6)
(2,4)
(4,4)
(6,6)
(3,4)
(5,6)
The set of paths to destination 2 are shown below:
1
2
1
3
6
5
2
3
4
1
2
3
2
4
5
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Now continue the algorithm after the link between node 2 and 4 goes down.
Iteration
Before
break
1
2
3
4
Node 1
(2,3)
Node 3
(4,3)
Node 4
(2,1)
Node 5
(2,4)
Node 6
(3,4)
(2,3)
(3,5)
(4,6)
(2,3)
(3,5)
( 4 , 10)
(2,3)
(3,7)
( 4 , 10 )
(2,3)
(3,7)
( 4 , 12 )
(1,5)
(4,3)
(6,5)
(1,5)
(4,7)
(6,5)
(1,5)
(4,7)
(6,5)
(1,5)
(4,9)
(6,7)
(1,8)
( 3, 5 )
(5,7)
(1,8)
(3,5)
(5,7)
(1,8)
(3,7)
(5,7)
(1,8)
(3,7)
(5,7)
(2,4)
(4,4)
(6,6)
(2,4)
(4,8)
(6,6)
(2,4)
(4,8)
(6,6)
(2,4)
( 4 , 10 )
(6,8)
(3,4)
(5,6)
(3,4)
(5,6)
(3,6)
(5,6)
(3,6)
(5,6)
The new set of paths are shown below:
1
2
1
3
6
5
2
3
4
1
2
3
2
4
5
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Chapter 8 Solutions
Solutions to Chapter 8
8.1. The IP header checksum only verifies the integrity of IP header. Discuss the pros and cons of doing the
checksum on the header part versus on the entire packet.
Solution:
Error checking in the header is more important because the packet is routed according to the header
information. In addition, the delivery of the data at the destination to the higher layers also requires the
header information. Thus error checking of the header protects against misdelivery of the information.
Restricting the error checking to the header also simplifies the implementation in the nodes, requires
less checksum bits, and prevents unnecessary packet discard. Some higher layers can tolerate some
data errors, and higher layers also have the option of performing retransmission.
8.2. Identify the address class of the following IP addresses: 200.58.20.165; 128.167.23.20; 16.196.128.50;
50.156.10.10; 250.10.24.96.
Solution:
An IP address has a fixed length of 32 bits, where the most significant bits identify the particular
class. Therefore, to identify the address class we need to convert the dotted-decimal notation back
into its binary counterpart, and compare the binary notation to the class prefixes shown in Figure
8.5 in the text. (Recall that the dotted-decimal notation was devised to communicate addresses
more readily to other people. In this notation, the 32 bits are divided into four groups of 8 bits –
separated by periods – and then converted to their decimal counterpart.) The first few bits (shown
in red) of the address can be used to determine the class.
2 7 26
128 64
25
32
24
16
23
8
22
4
21
2
20
1
200.58.20.165
11001000.00111010.00010100.10100101
Class C
128.167.23.20
10000000.10100111.00010111.00010100
Class B
16.196.128.50
00010000.11000100.10000000.00110010
Class A
150.156.10.10
10010110.10011100.00001010.00001010
Class B
250.10.24.96
11111010.00001010.00011000.01100000
Class E
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8.6. A host in an organization has an IP address 150.32.64.34 and a subnet mask 255.255.240.0. What is the
address of this subnet? What is the range of IP addresses that a host can have on this subnet?
Solution:
Address:
10010110 00100000 01000000 00100010
Mask:
11111111 11111111 11110000 00000000
Subnet:
10010110 00100000 01000000 00000000
Host:
From: 10010110 00100000 01000000 00000001
To: 10010110 00100000 01001111 11111110
8.7. A university has 150 LANs with 100 hosts in each LAN.
Solutions follow questions:
(a) Suppose the university has one Class B address. Design an appropriate subnet addressing scheme.
A Class B address has 14 bits for the network ID and 16 bits for the host ID. To design an
appropriate subnet addressing scheme we need to decide how many bits to allocate to the host ID
versus the subnet ID. We can choose either 7 bits or 8 bits to identify the hosts.
If we allocate 8 bits for to identify the host, as shown below, then there are sufficient subnet-id bits
to cover up to 28=256 LANs and enough host-id bits to cover up to 256 hosts for each LAN. The
subnet mask in this case is 255.255.255.0
1 0
0
1
Network-id
Subnet-id
15 16
Host-id
23 24
31
Subnet mask: 255.255.255.0
If we allocate 7 bits for to identify the host, as shown below, then there are sufficient subnet-id bits
to cover up to 29=512 LANs and enough host-id bits to cover up to 128 hosts for each LAN. The
subnet mask in this case is 255.255.255.128.
The choice between 7 or 8 bits to represent the hosts depends on which is likely to grow more, the
number of subnets or the number of hosts in a LAN. Alternatively a variable-length prefix scheme
using 7-bit host addresses, and grouping these form larger subnets provides greater flexibility in
accommodating future changes.
(b) Design an appropriate CIDR addressing scheme.
CIDR addressing scheme involves devising a prefix length that indicates the length of the network
mask. In this case, 8 bits are required to identify each LAN (since 127 < 150 < 255) and 7 bits are
required to identify each host in each LAN (since 63 < 100 < 127). Therefore a CIDR address
would use a 17-bit prefix, and thus have an address of the form address/17.
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8.9. A packet with IP address 150.100.12.55 arrives at router R1 in Figure 8.8. Explain how the packet is
delivered to the appropriate host.
Solution:
The packet with IP address 150.100.12.55 arrives from the outside network. R1 has to know the
next-hop router or host to send the packet to. The address corresponds to the binary string
10010110.01100100.00001100.00110111. R1 knows that a 9 bit subnet field is in use so it applies
the following mask to extract the subnetwork address from the IP address.
11111111.11111111.11111111.10000000
The resulting IP address is 10010110.01100100.00001100.00000000 and corresponds
150.100.12.0. This indicates that the host is in subnet 150.100.12.0, so the router transmits the IP
packet on this (attached) LAN.
8.15. Suppose four major ISPs were to emerge with points of presence in every major region of the world. How
should a CIDR scheme treat these ISPs in relation to addressing for each major region?
Solution:
The networks of these ISPs will span across
national and geographical boundaries and
connected in a non-hierarchical manner. These large global ISPs constitute major transit routing
domains, so it makes sense to assign them blocks of unique IP addresses and to require that
domains attached to them should begin with the transit domain’s prefix. The blocks of addresses
should be managed so that fragmentation of the CIDR block does not take place. For example,
when a customer of the ISP changes to another ISP, the customer is required to return the block of
addresses. This policy enables CIDR to be effective in controlling the size of routing tables.
8.16. Discuss the difficulties with using actual time in the TTL field.
Solution:
Unlike the number of hops, which is predictable if a packet is routed correctly, the actual time that it
takes to go through the route is not predictable. Therefore the amount of time that a packet stays in the
network is not necessarily an indication of misrouting. To allow an upper limit for delay across the
network TTL field would become a very large number. It is also more complex to track and update the
TTL according to actual time as the packet traverses the network.
8.18. Suppose a router receives an IP packet containing 600 data bytes and has to forward the packet to a network
with maximum transmission unit of 200 bytes. Assume that the IP header is 20 bytes long. Show the fragments
that the router creates and specify the relevant values in each fragment header (i.e., total length, fragment offset,
and more bit).
Solution:
Given:
IP packet = 600 data bytes
MTU = 200 bytes
IP header = 20 header bytes
Maximum possible data length per fragment = MTU – IP header = 200 – 20 = 180 bytes.
The data length of each fragment must be a multiple of eight bytes; therefore the maximum
number of data bytes that can be carried per fragment is 22*8=176.
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The data packet must be divided into 4 frames, as shown by the following calculations:
176 + 176 + 176 + 72 = 600
20 + 20 + 20 + 20
196 196 196 92
The sequence of frames and packet headers is shown below:
Total length
Id
Mf
Fragment Offset
Original Packet
620
x
0
0
Fragment 1
196
x
1
0
Fragment 2
196
x
1
22
Fragment 3
196
x
1
44
Fragment 4
92
x
0
66
8.34. Consider the three-way handshake in TCP connection setup.
Solutions follow questions:
(a) Suppose that an old SYN segment from station A arrives at station B, requesting a TCP connection.
Explain how the three-way handshake procedure ensures that the connection is rejected.
In a three-way handshake procedure, one must ensure the selection of the initial sequence number
is always unique. If station B receives an old SYN segment from A, B will acknowledge the request
based on the old sequence number. When A receives the acknowledgment segment from B, A will
find out that B received a wrong sequence number. A will discard the acknowledgment packet and
reset the connection.
(b) Now suppose that an old SYN segment from station A arrives at station B, followed a bit later by an old
ACK segment from A to a SYN segment from B. Is this connection request also rejected?
If an old SYN segment from A arrives at B, followed by an old ACK segment from A to a SYN
segment from B, the connection will also be rejected. Initially, when B receives an old SYN
segment, B will send a SYN segment with its own distinct sequence number set by itself. If B
receives the old ACK from A, B will notify A that the connection is invalid since the old ACK
sequence number does not match the sequence number previously defined by B. Therefore, the
connection is rejected.
8.44. What is the maximum width of a RIP network?
Solution:
The maximum width of an RIP network is 15 nodes. Since the maximum number of hops in RIP
network is 15, node 16 represents infinity.
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8.45. Let’s consider the bandwidth consumption of the RIP protocol.
Solutions follow questions:
(a) Estimate the number of messages exchanged per unit time by RIP.
A router implementing RIP sends an update message every 30 seconds. Assume that the typical
node has D neighbors and that the number of nodes in the network is N. The number of message
exchanges is then ND/30 messages per second.
(b) Estimate the size of the messages exchanged as a function of the size of the RIP network.
A RIP message consists of a four-byte header plus 20-byte per entry and up to 25 entries per
message. Therefore, each node will send out a message of at most 20(N 1) + 4 bytes every 30
seconds.
(c) Estimate the bandwidth consumption of a RIP network.
The bandwidth consumption is at most [ ND [20 (N 1) + 4]] / 30 bytes per second.
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Solutions to Chapter 10
10.1. Can all traffic in an IntServ router be of guaranteed service type? Explain.
Solution:
All the traffic can be of guaranteed service only if all the flows are shaped and if they strictly comply
with the traffic specifications. However in practice the total amount of traffic that can be admitted will
have some limit. The reason is that to bound the delay, the maximum queue length in the router must
be below a certain threshold. If all of the flows are well behaved and without bursts, the queue size
will always be reasonably small. However if the flows are bursty, the queue may grow to levels that
can cause large unacceptable delays. To guarantee the delay bounds, extra bandwidth should be
assigned based to accommodate the burstiness of the flows. This overbooking can sometimes lead to
the queue becoming empty, and hence result in underutilization of the output link. Thus, it is
reasonable to fill the output bandwidth during such periods with best effort traffic. So in summary, to
efficiently utilize the bandwidth and achieve higher throughputs it is necessary to admit best effort
traffic alongside guaranteed service traffic.
10.2. Develop a packet-scheduling approach to provide guaranteed service, controlled load service, and best effort
service in a router. Can any mix of these three traffic types be supported? Explain.
Solution:
Separate queues must be provided for the packets from each type of service in order for their packets
to be serviced appropriately. The set of queues belonging to guaranteed service should always
receive service with a rate R, which should be larger than the total guaranteed service traffic. The
aggregate rate R is guaranteed and controlled by giving priority to the guaranteed service queues and
controlling the service they receive by a leaky bucket shaper with a sustained rate of R. Individual
queues within the guaranteed service group are serviced in a round-robin fashion while for each
queue its service is controlled by a dedicated leaky bucket shaper with the parameters set based on
the requirements of the flow assigned to that queue.
The set of controlled load service queues are serviced during the periods that guaranteed service
queues are not serviced. The capacity in excess of R is available to controlled load traffic. Flows are
assigned to different queues and weighted fair queueing can be used within the group to service the
queues depending on their requirements.
Best effort traffic is given the lowest priority. The queue is serviced only if the other two sets of
queues are empty.
10.3. RSVP signaling is very different from ATM signaling. Discuss the differences and list the advantages and
disadvantages of each protocol.
Solution:
RSVP uses Receiver-Initiated Reservation rather than ATM’s per-hop path request to reserve
network resources. RSVP is intended for use with multicasting and is receiver oriented. ATM is
designed for unicast traffic and is sender oriented.
RSVP receiver-initiated reservation
Advantage: interoperates with non-RSVP routers; soft state provides robustness with respect to
faults, requests can be merged reducing the reservation load on the routers.
Disadvantage: soft state – requires periodic updates, receiver needs to know the path for connection.
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ATM per-hop path request
Advantage: hard state – reserve only once.
Disadvantage: Individual VCC is required for each host in multicast application. Source requests the
connections so it needs to know the requirements of the receiver.
10.4. Consider the merging of reservations in RSVP. Suppose that there is a single sender and 2n receivers that are
reached through a path that consists of a depth n binary tree that has a router located at each node. What is the total
number of reservation request messages generated by the receivers?
Solution:
The total number of reservation request messages generated by the receivers and routers are
equivalent to finding the branches of a binary tree with a depth of n. Let Nn be the number of
branches in a binary tree with depth n. At level n, there 2n new branches, and so we have the
recursion Nn = Nn–1 + 2n. (Draw a picture of a tree to verify this.) It is easy to then verify that Nn, total
number of request messages, is 2n+1 – 1.
10.5. Suppose RSVP is used to set up a virtual private network to interconnect n users across an Internet. Discuss
the appropriateness of the three RSVP reservation styles for this application.
RSVP has three reservation styles, Wildcard Filter, Fixed Filter and Shared Explicit.
Wild card filter (WF) creates a single reservation shared by all senders. It is designed for applications
that are unlikely to have multiple senders transmitting simultaneously. It is not appropriate for VPN
applications because WF cannot provide the receiver requests if more than one sender is transmitting
at the same time.
Fixed filter (FF) creates a distinct reservation for each sender. It is an appropriate solution for VPN
requirements. Each sender (VPN) has its own resource reservation and this resource is shared by
number of receivers (hosts located in different location belong to the same VPN).
Shared explicit (SE) creates a single reservation shared by a set of explicit senders. It may also be
appropriate in VPN applications under certain conditions. The merging of the reservation between
senders can result in a more efficient use of network resources.
10.6. Select and justify a choice an RSVP reservation style that is appropriate for the following situations:
Solutions follow questions:
(a) A multiuser application where a video stream from each participant must be delivered to every other
participant.
This situation is a multipoint to multipoint application so the Fixed Filter (FF) is the appropriate choice.
FF creates a distinct reservation for each participant (sender) and it can merge downstream requests
and forward the sum of flow spec upstream. It is an appropriate reservation style because each
participant (sender) will only receive a single reservation request that has been merged by a multiple
of other participants. This simplifies the processing overhead for each sender and allows the
processing power to focus on accepting multiple video streams for other participants.
(b) An audioconferencing application that is self-limiting in that more than two participants are unlikely to
speak simultaneously.
This audio conferencing application is best suited with the Wild card Filter (WF) style. The WF-style
is suitable for applications that are unlikely to have multiple senders transmitting simultaneously. In
this application the self-limiting feature prevents two participants from speaking simultaneously.
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(c) A multiuser application in which each participant must receive explicit permission before transmitting
information.
The Shared Explicit (SE) reservation style is best suited for this multiuser application. The SE style
creates a single reservation shared by a set of explicit senders. In this application, only the
participant that receives an explicit permission is allowed to join a single reservation. Any participant
that does not receive permission is not allowed to use this single reserved resource.
10.7. What information needs to be carried by Resv messages in RSVP to set up guaranteed IP service with a
specific bandwidth and a given delay-bound requirement? How does a router process this information?
Solution:
The information that must be carried by Resv message in RSVP to set up guaranteed IP service with
a specific bandwidth and a given delay bound is: the common header, SESSION, RSVP_HOP,
TIME_VALUES, STYLE, and Fixed Filter flow descriptor along with flow spec and filter spec. When a
router receives this Resv message, it reserves the necessary resources based on the flow
specification, which in turn governs the bandwidth and delay-bound requirements. Next, the router
exams the filter spec to determine the data packets that correspond to the defined flow spec.
10.8. Discuss the interplay between a QoS routing algorithm, RSVP, and integrated services IP.
Solution:
Integrated services IP provides QoS packet delivery between the source and destination pair. The
QoS parameters are dependent on the source requirements, for example, bandwidth requirement and
delay-bound requirement. To provide integrated services, it is necessary for the routing protocol
(network layer) to provide QoS routing. QoS routing provides for the selection of paths with specific
attributes that can support specified QoS requirements. RSVP provides the means for reserving
resources along paths identified by QoS routing.
10.9. Explain how the soft-state feature of RSVP allows it to adapt to failures in the network.
Solution:
The RSVP soft-state approach requires each node to refresh its reservation periodically by using
Path and Resv messages. When a state is not refreshed within a certain time-out period, the state is
deleted. Upon a link failure within the network, a PathErr message is sent to the ingress node
(source node). The ingress node simply defines a new route by refreshing the Path message to a
different next hop. The old failure path will eventually tear down after the timer is expired. This
approach provides error recovery upon failure, however, it is necessary for the ingress node to keep
updating the status of the link-state topology and determine the next possible route in case of the
primary route fails.
10.10. Discuss whether RSVP provides a scalable means to set up connections to a session that consists of the
broadcasting of a given TV program.
Solution:
Broadcasting a TV program is a point to multipoint multicast application. RSVP does provide a
scalable means to set up connections to a session in multicast application. RSVP performs the
resource reservation for multicast applications, adapting dynamically to changing group membership
and changing route. Also, RSVP requires the receiver to initiate and maintain the resource
reservation rather than obtaining the bandwidth requirement from each receiver. This process avoids
an implosion problem and minimizes the processing overhead to the sender. Finally, the RSVP
routers provide different reservation styles and the requests can merge in several ways according to
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the reservation style. This prevents duplicate resource reservations if multiple downstream receivers
belong to the same multicast group.
10.11. Explain what happens if RSVP is used to set up small-bandwidth flows on a large-bandwidth link?
well does RSVP scale in this situation?
How
Solution:
If RSVP is used to set up small-bandwidth flows on a large-bandwidth link, the router will have to
update the available resource (bandwidth) every time there are small-bandwidth flow request for
resources. Moreover, the router needs to keep track of the available resources, and reject resource
reservation requests when the available bandwidth cannot support a particular requirement. The
router has to keep track of every single flow resource usage during the setup and tear down
procedure. All of these activities consume a lot of processing power when the link has large
bandwidth being shared by large numbers of small-bandwidth flows.
10.12. Obtain a packet capture file for RSVP either in the lab or downloaded from a web site. Examine the details
of Path, Resv, and refresh messages.
Solution:
The following text shows the details of the first PATH message for MPLS path setup shown in Figure
2.27. The file is obtained by printing the capture data from Ethereal. We highlight in color the main fields
and interesting information. In particular note the explicit route identified in the message and the LABEL
REQUEST. The TIME VALUES field is set to 30000 ms, that is, 30 second refresh period.
Frame 1 (306 bytes on wire, 306 bytes captured)
Arrival Time: Feb 10, 2000 08:49:03.806994000
Time delta from previous packet: 0.000000000 seconds
Time relative to first packet: 0.000000000 seconds
Frame Number: 1
Packet Length: 306 bytes
Capture Length: 306 bytes
Ethernet II, Src: 00:90:92:9d:94:01, Dst: 00:d0:63:c3:b8:47
Internet Protocol, Src Addr: 17.3.3.3 (17.3.3.3), Dst Addr: 16.2.2.2 (16.2.2.2)
Resource ReserVation Protocol (RSVP): PATH Message. SESSION: IPv4-LSP, Destination 16.2.2.2, Tunnel ID 1, Ext ID
11030303. SENDER TEMPLATE: IPv4-LSP, Tunnel Source: 17.3.3.3, LSP ID: 1.
RSVP Header. PATH Message.
RSVP Version: 1
Flags: 00
Message Type: PATH Message. (1)
Message Checksum: 0xdb58 (correct)
Sending TTL: 254
Message length: 264
SESSION: IPv4-LSP, Destination 16.2.2.2, Tunnel ID 1, Ext ID 11030303.
Length: 16
Class number: 1 - SESSION object
C-type: 7 - IPv4 LSP
Destination address: 16.2.2.2 (16.2.2.2)
Tunnel ID: 1
Extended Tunnel ID: 285410051 (17.3.3.3)
HOP: IPv4, 210.0.0.1
Length: 12
Class number: 3 - HOP object
C-type: 1 - IPv4
Neighbor address: 210.0.0.1
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Logical interface: 0
TIME VALUES: 30000 ms
Length: 8
Class number: 5 - TIME VALUES object
C-type: 1
Refresh interval: 30000 ms (30 seconds)
EXPLICIT ROUTE: IPv4 210.0.0.2, IPv4 204.0.0.1, IPv4 207.0.0.1, ...
Length: 60
Class number: 20 - EXPLICIT ROUTE object
C-type: 1
IPv4 Subobject - 210.0.0.2, Strict
Strict Hop
Type: 1 (IPv4)
Length: 8
IPv4 hop: 210.0.0.2
Prefix length: 32
IPv4 Subobject - 204.0.0.1, Strict
Strict Hop
Type: 1 (IPv4)
Length: 8
IPv4 hop: 204.0.0.1
Prefix length: 32
IPv4 Subobject - 207.0.0.1, Strict
Strict Hop
Type: 1 (IPv4)
Length: 8
IPv4 hop: 207.0.0.1
Prefix length: 32
IPv4 Subobject - 202.0.0.1, Strict
Strict Hop
Type: 1 (IPv4)
Length: 8
IPv4 hop: 202.0.0.1
Prefix length: 32
IPv4 Subobject - 201.0.0.1, Strict
Strict Hop
Type: 1 (IPv4)
Length: 8
IPv4 hop: 201.0.0.1
Prefix length: 32
IPv4 Subobject - 200.0.0.1, Strict
Strict Hop
Type: 1 (IPv4)
Length: 8
IPv4 hop: 200.0.0.1
Prefix length: 32
IPv4 Subobject - 16.2.2.2, Strict
Strict Hop
Type: 1 (IPv4)
Length: 8
IPv4 hop: 16.2.2.2
Prefix length: 32
LABEL REQUEST: Basic: L3PID: IP (0x0800)
Length: 8
Class number: 19 - LABEL REQUEST object
C-type: 1
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L3PID: IP (0x0800)
SESSION ATTRIBUTE: SetupPrio 0, HoldPrio 0, SE Style, [sys17-3_t1]
Length: 20
Class number: 207 - SESSION ATTRIBUTE object
C-type: 7 - IPv4 LSP
Setup priority: 0
Hold priority: 0
Flags: 0x04
.... ...0 = Local protection not desired
.... ..0. = Label recording not desired
.... .1.. = SE style desired
.... 0... = Bandwidth protection not desired
...0 .... = Node protection not desired
Name length: 10
Name: sys17-3_t1
SENDER TEMPLATE: IPv4-LSP, Tunnel Source: 17.3.3.3, LSP ID: 1.
Length: 12
Class number: 11 - SENDER TEMPLATE object
C-type: 7 - IPv4 LSP
Sender IPv4 address: 17.3.3.3 (17.3.3.3)
Sender LSP ID: 1
SENDER TSPEC: IntServ: Token Bucket, 625000 bytes/sec.
Length: 36
Class number: 12 - SENDER TSPEC object
C-type: 1 - Integrated Services
Message format version: 0
Data length: 7 words, not including header
Service header: 1 - Traffic specification
Length of service 1 data: 6 words, not including header
Token Bucket TSpec: Rate=625000 Burst=1000 Peak=625000 m=0 M=0
Parameter 127 - Token bucket
Parameter 127 flags: 0x00
Parameter 127 data length: 5 words, not including header
Token bucket rate: 625000
Token bucket size: 1000
Peak data rate: 625000
Minimum policed unit [m]: 0
Maximum packet size [M]: 0
ADSPEC
Length: 84
Class number: 13 - ADSPEC object
C-type: 2
Message format version: 0
Data length: 19 words, not including header
Default General Parameters
Service header 1 - Default General Parameters
Break bit not set
Data length: 8 words, not including header
IS Hop Count - 1 (type 4, length 1)
Path b/w estimate - 1250000 (type 6, length 1)
Minimum path latency - 0 (type 8, length 1)
Composed MTU - 1500 (type 10, length 1)
Guaranteed Rate
Service header 2 - Guaranteed Rate
Break bit not set
Data length: 8 words, not including header
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End-to-end composed value for C - 169500 (type 133, length 1)
End-to-end composed value for D - 1200 (type 134, length 1)
Since-last-reshaping point composed C - 169500 (type 135, length 1)
Since-last-reshaping point composed D - 1200 (type 136, length 1)
Controlled Load
Service header 5 - Controlled Load
Break bit not set
Data length: 0 words, not including header
The following text shows the details of the first RESV message in the opposite direction for the MPLS
path setup shown in Figure 10.27. In particular note the last entry which assigns the label 16 to the path.
Frame 2 (146 bytes on wire, 146 bytes captured)
Arrival Time: Feb 10, 2000 08:49:03.909463000
Time delta from previous packet: 0.102469000 seconds
Time relative to first packet: 0.102469000 seconds
Frame Number: 2
Packet Length: 146 bytes
Capture Length: 146 bytes
Ethernet II, Src: 00:d0:63:c3:b8:47, Dst: 00:90:92:9d:94:01
Internet Protocol, Src Addr: 210.0.0.2 (210.0.0.2), Dst Addr: 210.0.0.1 (210.0.0.1)
Resource ReserVation Protocol (RSVP): RESV Message. SESSION: IPv4-LSP, Destination 16.2.2.2, Tunnel ID 1, Ext ID
11030303. FILTERSPEC: IPv4-LSP, Tunnel Source: 17.3.3.3, LSP ID: 1.
RSVP Header. RESV Message.
RSVP Version: 1
Flags: 00
Message Type: RESV Message. (2)
Message Checksum: 0x130b (correct)
Sending TTL: 255
Message length: 108
SESSION: IPv4-LSP, Destination 16.2.2.2, Tunnel ID 1, Ext ID 11030303.
Length: 16
Class number: 1 - SESSION object
C-type: 7 - IPv4 LSP
Destination address: 16.2.2.2 (16.2.2.2)
Tunnel ID: 1
Extended Tunnel ID: 285410051 (17.3.3.3)
HOP: IPv4, 210.0.0.2
Length: 12
Class number: 3 - HOP object
C-type: 1 - IPv4
Neighbor address: 210.0.0.2
Logical interface: 0
TIME VALUES: 30000 ms
Length: 8
Class number: 5 - TIME VALUES object
C-type: 1
Refresh interval: 30000 ms (30 seconds)
STYLE: Shared-Explicit (18)
Length: 8
Class number: 8 - STYLE object
C-type: 1
Flags: 0x00
Style: 0x000012 - Shared-Explicit
FLOWSPEC: Controlled Load: Token Bucket, 625000 bytes/sec.
Length: 36
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Class number: 9 - FLOWSPEC object
C-type: 2
Message format version: 0
Data length: 7 words, not including header
Service header: 5 - Controlled Load
Length of service 5 data: 6 words, not including header
Token Bucket: Rate=625000 Burst=1000 Peak=1.#INF m=0 M=0
Parameter 127 - Token bucket
Parameter 127 flags: 0x00
Parameter 127 data length: 5 words, not including header
Token bucket rate: 625000
Token bucket size: 1000
Peak data rate: 1.#INF
Minimum policed unit [m]: 0
Maximum packet size [M]: 0
FILTERSPEC: IPv4-LSP, Tunnel Source: 17.3.3.3, LSP ID: 1.
Length: 12
Class number: 10 - FILTER SPEC object
C-type: 7 - IPv4 LSP
Sender IPv4 address: 17.3.3.3 (17.3.3.3)
Sender LSP ID: 1
LABEL: 16
Length: 8
Class number: 16 - LABEL object
C-type: 1 (Packet Label)
Label: 16
The following text shows the details of the first refresh message sent about 23 seconds after the first
frame. The contents of the RSVP message are identical to the previous RESV message.
Frame 3 (146 bytes on wire, 146 bytes captured)
Ethernet II, Src: 00:d0:63:c3:b8:47, Dst: 00:90:92:9d:94:01
Internet Protocol, Src Addr: 210.0.0.2 (210.0.0.2), Dst Addr: 210.0.0.1 (210.0.0.1)
Resource ReserVation Protocol (RSVP): RESV Message. SESSION: IPv4-LSP, Destination 16.2.2.2, Tunnel ID 1, Ext ID
11030303. FILTERSPEC: IPv4-LSP, Tunnel Source: 17.3.3.3, LSP ID: 1.
RSVP Header. RESV Message.
RSVP Version: 1
Flags: 00
Message Type: RESV Message. (2)
Message Checksum: 0x130b (correct)
Sending TTL: 255
Message length: 108
SESSION: IPv4-LSP, Destination 16.2.2.2, Tunnel ID 1, Ext ID 11030303.
Length: 16
Class number: 1 - SESSION object
C-type: 7 - IPv4 LSP
Destination address: 16.2.2.2 (16.2.2.2)
Tunnel ID: 1
Extended Tunnel ID: 285410051 (17.3.3.3)
HOP: IPv4, 210.0.0.2
Length: 12
Class number: 3 - HOP object
C-type: 1 - IPv4
Neighbor address: 210.0.0.2
Logical interface: 0
TIME VALUES: 30000 ms
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Length: 8
Class number: 5 - TIME VALUES object
C-type: 1
Refresh interval: 30000 ms (30 seconds)
STYLE: Shared-Explicit (18)
Length: 8
Class number: 8 - STYLE object
C-type: 1
Flags: 0x00
Style: 0x000012 - Shared-Explicit
FLOWSPEC: Controlled Load: Token Bucket, 625000 bytes/sec.
Length: 36
Class number: 9 - FLOWSPEC object
C-type: 2
Message format version: 0
Data length: 7 words, not including header
Service header: 5 - Controlled Load
Length of service 5 data: 6 words, not including header
Token Bucket: Rate=625000 Burst=1000 Peak=1.#INF m=0 M=0
Parameter 127 - Token bucket
Parameter 127 flags: 0x00
Parameter 127 data length: 5 words, not including header
Token bucket rate: 625000
Token bucket size: 1000
Peak data rate: 1.#INF
Minimum policed unit [m]: 0
Maximum packet size [M]: 0
FILTERSPEC: IPv4-LSP, Tunnel Source: 17.3.3.3, LSP ID: 1.
Length: 12
Class number: 10 - FILTER SPEC object
C-type: 7 - IPv4 LSP
Sender IPv4 address: 17.3.3.3 (17.3.3.3)
Sender LSP ID: 1
LABEL: 16
Length: 8
Class number: 16 - LABEL object
C-type: 1 (Packet Label)
Label: 16
10.13. What RSVP reservation style is appropriate for distributed network gaming?
Solution:
Distributed network gaming is similar to video conferencing in nature. The signals from each of the
participants are received by all of the other participants. Therefore the fixed-filter style is appropriate
in this case. The shared-explicit style may also be appropriate if the number of simultaneous signals
is not too large.
10.14. What RSVP reservation style is appropriate for instant messaging? For stock price notification?
Solution:
For instant messaging the fixed-filter style is appropriate because the messages are usually
exchanged between the two participants.
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For stock price notification the wildcard-filter style is appropriate because the notification can be
received by any number of receivers similar to a multicast session. A customized notification however
could be different and may require fixed-filter style for individual notifications.
10.15. A sender transmits layered video information. Provide appropriate reservation messages for the receivers in
Figure 10.5 if Rx1 is on a 802.11 wireless LAN, Rx2 is on a 3G cell phone, and Rx3 is a workstation on a fast
Ethernet LAN.
Solution:
Rx1 sends a reservation request message (Resv) for a shared–explicit style connection in which the
flowspec indicates a flow rate of 2 Mbps. The request can be represented by SE(Rx1,{2 Mbps}).
Similarly Rx2 sends the request message SE(Rx2,{144 Kbps}) and Rx3 sends the request message
SE(Rx3,{100 Mbps}.
10.16. Derive the state lifetime L in RSVP.
Solution:
A local node should be able to tolerate losses of the refresh signal as long as at least one out of K
consecutive messages gets through. In the worst case this message could be the last one in a
window of K messages. Therefore L must be larger than K times the refresh period. The refresh
period is randomized number with a uniform distribution in the range of [0.5R, 1.5R]. In the worst case
the refresh period could be 1.5R. As a result to tolerate the worst-case scenario, L must be larger
than K*1.5*R.
10.17. Discuss possible scheduling mechanisms (e.g., head-of-line priority and weighted fair queueing) that you
could use to implement EF PHB, AF PHB group, and DE PHB. Discuss the trade-offs between performance and
complexity.
Solution:
If head-of-line (HOL) priority is used to implement EF PHB, AFPHB and DE PHB, at least three
separate queues will be defined with different priority classes. The EF PHB assigns the highest
priority followed by AF PHB. DE PHB has the lowest priority. HOL priority is easy to implement to
provide different levels of service to different classes (PHB). However, HOL priority does not allow for
providing some degree of guaranteed access to the transmission bandwidth for the lower priority
classes. Another problem is that it does not discriminate users of the same priority. A fairness
problem arises when certain users hog the bandwidth by sending an excessive number of packets.
In general, HOL priority may be able to provide EF PHB service requirements. On the other hand, it
is necessary to have a large buffer queue for AF PHB to achieve low packet drop rate due to the
fairness problems. As a result, HOL is not the most suitable traffic management for different PHB
applications.
In weighted fair queuing each PHB still has its own queue, but each PHB also has a weight to
determine the relative service share it receives. Therefore, EF PHB gets a somewhat larger share of
the bandwidth to enable it to meet its delay requirements; and DE PHB only some minimum share of
the bandwidth. Weighted fair queuing is a more complicated traffic management approach than the
HOL. However, it can prevent the fairness problem. Weighted fair queuing ensures both EF and AF
PHB to have a fair share amount of bandwidth to provide high assurance while maintaining EF to
have low latency and assured bandwidth. Complex policy control is necessary to determine the right
amount of bandwidth for each PHB.
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A practical approach to scheduling involves a combination of the above mechanisms. EF PHB traffic
is subjected to shaping and then placed in a queue that receives priority over all other traffic. The
shaping places a limit on the sustained rate, and hence assures other traffic of a certain amount of
bandwidth. AF and DE PHB traffic are subjected to weighted fair queueing in order to assure fairness
to the remaining bandwidth.
10.18. Perform a web search of router equipment that implements differentiated services IP. Find out what PHBs
are supported by the routers, and if possible, find out what scheduling mechanisms are used to implement the PHBs.
Solution (Based on CISCO 7400 series (7401):
Routers that implement differentiated services support at least real-time (EF PHB) and non real-time
forwarding, which are either only best effort or both best effort and guaranteed (AF PHB). Multiple
classes within the AF PHB are also supported in these routers.
The implementation of differentiated services and support for different PHBs is achieved in these
routers through a combination of different coordinated mechanisms:
•
Classification engines, which classify the applications according to differentiated services
specifications at the source
•
Class-based traffic shaping and bandwidth policing mechanisms
•
Class-based buffer management and congestion control mechanisms such as class-based
weighted random early detection (CBWRED)
•
Class-based scheduling such as class-based weighted fair queueing (CBWFQ)
On top of all of these mechanisms the router OS software is the main element that enables the router
to use the mentioned mechanisms to support the differentiated services.
10.19. Perform a web search for microelectronic chips that implement packet scheduling. What QoS mechanisms
are supported by these chips? At what bit rates? Do the chips require external memory and if so how much?
Explain the memory requirements.
Solution (Based on Vitesse VSC871, VSC881, Terachip TCF16x10, Agere P1-40SAX, BCM-832,
TSX-1200, TSX-1400):
Packet scheduling chips are designed either to be used in line cards or jointly with the switching
fabric. The chips usually come as part of a chip set. The scheduling chips support QoS by providing
multiple buffering queues that can be grouped in different classes. A mechanism is usually provided
to schedule the service among the classes. Also within each class an internal mechanism is provided
to schedule different queues. This basic structure can be easily used to achieve different qualities of
service. This is done through the control software by assigning the traffics to appropriate queues and
by setting the parameters for the scheduling mechanisms appropriately.
Some scheduling chips support only two queues for real-time and non-real-time traffics. Other chips
support 4 queues for real-time and multiple classes of non-real-time traffic. There are also chips that
support a larger number of queues, such as 16 per class, and the two-level scheduling mechanism.
The scheduling mechanisms include strict priority to distinguish real-time traffic from best-effort in the
simplest case, strict priority in first level and round-robin among the non-real-time to achieve fairness,
weighted round-robin to provide differentiation in addition to fairness, and finally weighted fair
queueing.
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The bit rates vary from several mega bits per second to hundreds of mega bits per second depending
on the target type of the routers that can use the schedulers. For example chips intended for use in
the core routers support multiple 3.125 Gbps interfaces that can be combined in groups of four to
support 10 Gbps ports. The schedulers usually need external buffering for buffering and queue
management.
10.20. Provide arguments for and against the statement: “It is essential that Ethernet provide QoS.”
Solution:
Ethernet is a link layer protocol that is primarily used in local area networks. The local traffic in a LAN
can be exchanged with low delay and low loss probability. Using LAN switches, partitioning the LAN
into smaller clusters interconnected through switches or bridges, and also using higher speeds
(10/100/1000) keeps the delay low as the LAN scales and as the traffic increases. Requiring the
support for QoS in LAN environment while it is not a critical issue would make the equipment more
complex and more expensive. However, the application of Ethernet is not limited to LAN environment.
In larger networks such as access networks the distances are larger, the aggregate traffic is greater,
and the potential for congestion high. In order to use Ethernet in larger networks while guaranteeing
the quality of service, one can argue that it is essential that Ethernet provide QoS.
10.21. Can MPLS make use of DS information to determine the forwarding experienced by different traffic flows?
If yes, explain how. If no, explain why not.
Solution:
Yes. At the ingress MPLS router, the router examines the DS field of the IP packet and maps the
required service into the label. The MPLS header has the EXP bits available for determining service
class. The label thus states the service level in the forwarding of the packet within the MPLS domain.
Within the MPLS domain, each router examines the label and performs the forwarding requirement.
At the egress node, the router pops the label and forwards the packet to the appropriate next hop.
Traffic shaping may be required at the egress node to ensure that traffic conforms to the policing
action in the subsequent network. The use of MPLS reduces the overhead required by a
conventional router to examine the packet header and determine the required service at each hop.
10.22. What is an appropriate traffic conditioner for EF PHB packets? AF PHB packets?
Solution:
EF PHB packets require low delay and low jitter treatment. These requirements are more readily and
efficiently met for traffic flows that are relatively smooth. Therefore EF flows should be subjected to
shaping that ensures that the packet flow is smooth flow. Packets that violate the smoothness
requirement should be discarded as they may end up not receiving the required low delay low jitter
treatment. AF flows are likely to be more bursty, and hence sustained rate and maximum allowable
bursting are more important. Packet buffering, marking, and dropping are appropriate traffic
conditioning mechanisms for AF flows.
10.23. Suppose that two RSVP-capable users are each connected to an intranet which in turn connects them through
an Internet backbone that provides differentiated services IP. Suppose that the intranets provide integrated services
IP. Explain the functions that must be implemented in the intranet routers in order to provide controlled-load service
and guaranteed service.
Solution:
The router in the intranet must be able to map the controlled-load service and guaranteed service into
the differentiated services domain. Since the controlled-load service is very similar to best effort
delivery with guaranteed resources, AF PHB is appropriate in the DS backbone. Therefore, if the flow
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conforms to the specification (service level agreement), the flow will experience low delay and low
loss. Guaranteed service requires bandwidth guarantee and low delay and jitter and hence should be
mapped to EF PHB.
10.24. Compare integrated services IP and differentiated services IP in terms of their ability to protect themselves
against theft of service.
Solution:
Integrated services IP provides better protection against theft of service than differentiated services
IP. Integrated services IP requires admission control before transporting any data packets. As a
result, each flow that is admitted should have enough network resources as long as each flow follows
its flow specification. If any flow does not conform and exceed its traffic specification, this particular
flow will be treated as best effort service to ensure fairness. To prevent theft of service, the
admission control process also needs to be secure. Differentiated services IP accords all packets with
the same DSCP the same PHB. Thus protection against theft of service involves controlling the
setting of DSCP values in the IP packet headers. This suggests strict control by edge routers at the
ingress to a DS domain.
10.25. Explain why shaping may be required at the egress node of a differentiated services domain.
Solution:
DS domains that belong to different administrations are likely to use traffic shape parameters in
negotiating the exchange of traffic between the domains. A shaper is required at the egress node to
ensure the traffic will conform to the policing action in the subsequent network. Therefore, a shaper is
important to ensure end-to-end QoS across multiple networks.
10.26. With reference to Figure 10.12, explain how the various types of routers are configured to support packet
transport services that require end-to-end low-delay and low-jitter, i.e. voice over IP.
Solution:
A packet transfer service that requires low delay and jitter should be assigned EF PHB. To ensure the
low delay and jitter at each DS router, the aggregate arrival rate of packets with EF PHB should be
less than the aggregate minimum allowed departure rate. Therefore every DS router must be
configured with a minimum departure rate for EF PHB independent of the other traffic. In addition, the
aggregate arrival rate must be shaped and policed so that it is always less than the minimum
configured departure rate. To ensure this requirement, the access DS router in the local DS domain
needs to support traffic classification and traffic conditioning. The interior DS router uses buffer
management and scheduling mechanisms to deliver the specific PHB according to the DS field of
each packet. The border DS router needs to shape the flow of packets such that it conforms to the
traffic conditioning agreement included in the SLA with the border DS router in the transit network.
The border DS router in the transit network uses traffic conditioning, including metering, marking and
shaping or dropping, to ensure that the arriving traffic behaves according to its agreement and to treat
the packets accordingly.
10.27. A common IP-over-ATM overlay model uses an ATM permanent virtual connection (PVC) to connect each
pair of IP routers thereby creating a fully meshed topology for the IP network. Thus N(N-1)/2 PVCs are needed if
there are N routers. Suppose that the routers run a link-state protocol (e.g., OSPF) among themselves. How many
advertisements the routers would receive if a given PVC is down. Repeat if a given router is down.
Solution:
If a given PVC is down, the two routers at the two ends detect the link failure and advertise it to all
other routers. This generates 2N(N – 1) advertisements. If a given router is down, each one of the
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other routers will detect the failure and advertise the failure of their links to the failed router to all other
routers. This will create (N – 1)(N – 1) advertisements.
10.28. Discuss the advantages and disadvantages of different label assignment approaches (i.e., traffic-driven,
topology-driven, and request-driven) in terms of number of connections and latency.
Solution:
Traffic-driven label assignment approach: In traffic driven label assignment the assignments follow
the traffic pattern in the network. The number of connections will depend on the traffic pattern and its
diversity. Monitoring the traffic pattern and adapting the connections will introduce latency.
Topology-driven label assignment approach: The assignment is slower as it may take time to collect
updated topology information. Traffic engineering can be incorporated in the assignment of
connections, which helps to optimize the resource utilization and increase the number of connections.
Request-driven label assignment approach: Request-driven label assignment can result in an efficient
use of the network resources because in this case one single node determines the routes that are
created. The node can optimize the route in terms of resource utilization. As a result the number of
connections that can be established can be maximized. The procedure requires however updated
state information from all nodes and processing, to optimize the routes that introduces higher latency.
10.29. MPOA uses flow detection to setup virtual circuits, whereas MPLS also relies on routing topology to
establish virtual circuits. Discuss the advantages and disadvantages of the two approaches.
Solution:
MPOA requires flow detection to set up a VCC
• Advantage: quick adaptivity to flow patterns
• Disadvantages: time delay on VCC establishment, processing load due to churn in VCC setup and
teardown
MPLS relies on routing topology to set up a VCC
• Advantage: less processing load than MPOA, traffic engineering capability
• Disadvantage: more complex and requires link state distribution or similar protocol to update the
network topology.
10.30. Consider explicit routing in MPLS.
Solutions follow questions:
(a) Can explicit routing in MPLS be used to select a route that meets the QoS requirements of a particular flow
(e.g., guaranteed bandwidth and delay)? Explain.
Yes, explicit routing in MPLS is capable of selecting a route to meet certain QoS requirements.
Explicit routing can establish a path between the source and destination that meets the QoS
requirements. However, it is necessary that the link-state distribution protocol be able to distribute
the required information.
(b) Give an example at a fine level of granularity that can make use of the QoS explicit routing in part (a).
Give another example at a coarser level of granularity.
An example at a fine level of granularity would be to define the QoS requirement on a per application
flow basis between a source and destination pair. An example at a coarse level of granularity would
be to define the QoS requirement between a set of network prefixes or between two network
domains.
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(c) Compare the case where the first LSR computes an explicit path to the case where all LSRs participate in
the selection of the path. What information does the first LSR require to select an explicit path?
If the first LSR computes an explicit path, the first LSR will need to have updated link-state
information for the whole MPLS domain. On the other hand, when all LSRs participate, each LSR
must have the signaling protocol for requesting, establishing, rejecting and tearing down the explicit
path. Moreover, in the latter case, each route does not need updated link-state topology of the MPLS
domain because each LSR can participate in the selection of path based on its currently known
network topology.
10.31. In an MPLS domain, rank the following three flows in terms of their level of aggregation: (a) all packets
destined to the same host; (b) all packets with the same egress router; (c) all packets with the same CIDR address.
Solution:
(a) All packets to the same host – This level of aggregation focuses only on the packets from any
source to the same destination by aggregating them into the same flow. This is a fine level of
traffic aggregation.
(b) All packets with the same egress router – All packets are aggregated into the same flow if they
are destined to the same egress router. This is a less specific level of traffic aggregation
compared to the previous case. More packets can aggregate into the flow as long as they are
destined to the same egress router. The egress router distributes the arriving packet to the
appropriate destination by examining each packet’s IP header.
(c) All packets with the same CIDR address – This is coarser level of flow aggregation. Basically all
packets destined to the same CIDR address (a number of network domains) will aggregate to the
same flow. A lot of packets can aggregate and the aggregation path is usually within the
backbone network.
10.32. Consider MPLS for unicast traffic. Explain what information is used to specify labels when labels are
determined according to (a) host pairs, (b) network pairs, (c) destination network, (d) egress router, (e) next-hop
AS, (f) destination AS.
Solution:
(a) Host pairs – the host pairs’ IP address
(b) Network pairs – the network pairs’ network prefix (IP prefix)
(c) Destination network – the network’s IP address and the AS number (if available)
(d) Egress router – the network or IP address that the egress router can reach.
(e) Next hop AS – the next hop AS number and the destination the next hop AS can reach
(f) Destination AS – the destination AS, or a network or IP prefix the AS contains
10.33. Explain why it may be useful for the MPLS header to carry a time-to-live value.
Solution:
A TTL field in an MPLS header can be useful for the same reasons it is used in IP: to terminate
packets caught in a loop; to assist utilities such as traceroute; to limit the reach of certain packets.
10.34. Compare the overall scalability of the following three cases:
Solutions follow questions:
(a) Layer 3 forwarding only: each router performs a longest-prefix match to determine the next forwarding
hop;
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Longest-prefix matching can take much processing time if the routing table is very large. This can
limit the rate at which packets can be routed.
(b) Layer 3 forwarding and some Layer 2 MPLS forwarding;
The layer 2 forwarding helps to reduce the load of the layer 3 forwarding. This method is very
scalable if the majority of packets use the layer 2 forwarding.
(c) Layer 2 MPLS forwarding only.
Layer 2 MPLS forwarding provides fast packet forwarding capabilities independent of the routing table
size. The packets will experience minimal delay on each hop. However the node cannot handle IP
packets that do not have MPLS labels.
10.35. Are MPLS label switching and packet filtering at firewalls compatible? Explain.
Solution:
Packet filtering at firewalls operates at or above the IP layer and MPLS label switching runs below IP
layer. Filtering requires extracting specific fields from a packet header and applying rules to the
values of the fields to determine the handling of a given packet. Network processors that can extract
and process multiple fields from multiple headers in a packet are now becoming available. In
particular MPLS is capable of establishing an explicit path with an egress node that has firewall
capability. The firewall at the egress node decides whether to forward or deny the packet that is
coming from an explicit path in the MPLS domain.
You may be interested in doing a web search for network processors. You will find Intel, IBM, and
many others involved in the development of network processors.
10.36. Discuss what factors determine the level of computational load for label assignment and distribution for the
following three cases: (a) topology-driven label assignment, (b) request-driven label assignment, (c) traffic-driven
label assignment.
Solution:
(a) Topology-driven label assignment – The label assignment for this approach basically follows the
shortest-path algorithms used in the IP network. As a result, label assignment is based on the
shortest path found by the IP layer and the computational load is minimal.
(b) Request-driven label assignment – The ingress router requests a traffic flow within an MPLS
domain. This requires each LSR to have the current status of the link-state network topology.
The computational load for label assignment and distribution is higher than for the topology-driven
approach.
(c) Traffic-driven label assignment – The router must monitor the traffic flow/pattern on each of its I/O
ports before assigning any label to the link. Computational load is the highest among the three
approaches because traffic monitoring requires extra processing overhead in this approach.
10.37. Compare MPLS label stacks and IP-over-IP encapsulation tunnels in terms of their traffic engineering
capabilities.
Solution:
Traffic engineering involves path pre-establishment and re-routing in case of failure. MPLS label
stacks are more flexible than IP-over-IP encapsulation in terms of traffic engineering. In MPLS label
stack, path re-routing simply changes the label value along the routing path without altering the
payload. However, in IP-over-IP encapsulation, the re-routing requires changing the IP address as
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well as manipulating the IP header (including header checksum and possibly fragmentation offset).
This results in complicated processing overhead for the router and poses delay on the packet
transmission.
10.38. Compare the following two approaches to traffic engineering:
Solutions follow questions:
(a) Offline computation: A central site knows the demand (total bandwidth requirement) between each pair of
LSRs and computes the paths in the network.
The centralized database contains an updated and accurate network state topology. As a result, it
provides the optimal path between any source and destination pair within the domain. However,
scalability is a problem. As the network becomes large, the volume of requests becomes difficult to
handle.
(b) Online computation: An ingress LSR independently computes a path to a specified egress LSR as new
demands arrive at the ingress LSR.
The processing load is distributed to individual ingress nodes. However, the ingress node must have
the updated link-state network topology and this topology may be less accurate compared with the
centralized database as multiple LSRs may assign paths at the same time.
10.39. Suppose that an MPLS network consisting of N routers supports C diffserv traffic classes. A trunk is defined
as traffic from a single traffic class that is aggregated into an LSP.
Solutions follow questions:
(a) What is the maximum number of trunks if a trunk is defined for every pair of ingress router and egress
router?
If all of the N routers are considered as ingress/egress pair routers, the the maximum number of
trunks for every pairs of ingress and egress router will be N(N − 1) × C.
(b) Suppose that the trunks in part (a) are merged if they have the same traffic class and the same egress
routers. What is the maximum number of multipoint-to-point trees rooted in an egress router?
The trunks are merged if they have the same traffic class and the same egress router, the maximum
number of trees rooted in an egress router is N × C.
(c) Can you think of other ways of merging trunks?
The trunks could be merged if they have the same egress router and even if they have a different
traffic class as long as enough network resources are allocated. The maximum number of trees
rooted in an egress router will become N.
10.40. Can an MPLS label correspond to all packets that match a particular integrated services filter specification?
If yes, explain how such a label may be used. If no, explain why not.
Solution:
The MPLS label is primarily concerned with the forwarding of a packet. The forwarding information is
accessed through the label in the MPLS header. The integrated services filter is concerned with
determining which packets are allowed access to a given set of reserved resources. However the
MPLS label can also be used to determine the type of scheduling that the associated packet flow is to
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undergo. Therefore the MPLS label can be used to determine that a certain packet flow is to undergo
a certain scheduling treatment that corresponds to the integrated service filter.
10.41. Suppose that network 2 in Figure 10.15 is a SONET network and that networks 1 and 3 are IP networks.
Explain the overlay and peer-to-peer models to network interconnection. What changes if network 1 and network 3
are MPLS networks? What changes if network 2 uses GMPLS?
Solution:
In the overlay model, the server network operates an independent control plane from the client
networks. The client networks receive the service from the server network in the form of logical links
connecting their border nodes. In this case the border nodes B and F contain packet-over-SONET
interfaces that connect to nodes C and E in the SONET network respectively. A SONET connection is
established between nodes C and E by the control plane in network 2 to provide the transport service
to the connection between networks 1 and 3.
In the peer-to-peer model, a single instance of the control plane is used for server and client
networks. In this case the nodes in the network 2 have an IP forwarding layer that interact with the IP
control plane in nodes in networks 1 and 3, providing a unified end-to-end IP control. The nodes in
network 2 translate the routing result of the IP layer into SONET connections to transport IP flows
between networks 1 and 3.
If networks 1 and 3 are MPLS networks, in both overlay and peer-to-peer cases the difference is in
how traffic flows are handled within the client networks. Instead of a larger IP network, we have a
larger combined MPLS network.
If network 2 supports GMPLS, then it is possible for GMPLS signaling to create a tunnel for the IP
connections between networks 1 and 3 with SONET connectivity across network 2.
10.42. Suppose that network 2 in Figure 10.15 provides optical wavelength service and that networks 1 and 3 are IP
networks. Explain the overlay and peer-to-peer models to network interconnection. What changes if network 1 and
network 3 are MPLS networks? What changes if network 2 uses GMPLS?
Solution:
In the overlay model, nodes B and F contain IP over λ interfaces that connect to nodes C and E in
network 2 respectively. Network 2 provides wavelength connectivity between nodes B and F. The
connectivity is established by the control plane in network 2 with the interaction with the control
planes in networks 1 and 3.
In the peer-to-peer model an IP forwarding layer on top of the wavelength service uses the IP
forwarding results to switch wavelengths across the network 2 to create connectivity between
networks 1 and 2.
If networks 1 and 3 are MPLS networks, instead of signaling for IP flows, MPLS connections are
mapped into wavelengths, a process which simplifies the signaling considerably.
If network 2 is GMPLS, it can simply create a tunnel for the IP connections between networks 1 and 3
using wavelength connectivity across the network 2, which simplifies the interconnection problem.
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10.43. Suppose that network 2 in Figure 10.15 provides optical wavelength services and that networks 1 and 3 are
SONET networks. Explain the overlay and peer-to-peer models to network interconnection. What changes if all the
networks use GMPLS?
Solution:
In overlay model, SONET connections are mapped into wavelengths in nodes B and F and
transported across network 2 using wavelength connectivity created by the control plane in network 2.
In peer-to-peer model the nodes in network 2 interact with SONET control plane to establish the endto-end SONET connections, which uses the wavelength connectivity in network 2 as part of its
transport path.
If all the networks use GMPLS, a unified control plane establishes the end-to-end connection, while
the connection is mapped on to the SONET transport in networks 1 and 3 and wavelength transport
in network 3.
10.44. Does the notion of label merging apply to GMPLS? Explain.
Solution:
The notion of label merging cannot be applied to GMPLS at all levels in a similar way it is applied to
MPLS. In the wavelength level for instance, multiple wavelengths cannot be merged into a new
wavelength.
10.45. Suppose that MPLS traffic runs over a SONET network. Is protection and restoration required at both the
SONET and MPLS levels? Explain.
Solution:
MPLS can provide both protection and restoration in an efficient way, as MPLS is the lowest level that
is aware of the link states while it has a comprehensive view of network topology and at the same
time has the means for traffic management. Protection at SONET level is done in the physical layer
and thus it is much faster, but requires more equipment deployment that is more costly. Restoration
relies on the network level routing protocols to calculate alternate paths. In general, MPLS does not
require SONET protection underneath, unless the service restoration times cannot be met by MPLS
mechanisms.
10.46. Consider a 4-node fully connected optical network. Suppose that each pair of nodes is connected by optical
fiber that carries 100 wavelengths in each direction. Compare the scale required to do OSPF routing with and
without link bundling.
Solution:
Without link bundling OSPF will distribute the link state for every wavelength as a separate routing
adjacency. With link bundling all parallel adjacencies can be aggregated into one bundled link
reducing the number of links and consequently the number of link state messages by a factor of 100.
10.47. Should RTP be a transport layer protocol? Explain why or why not.
Solution:
The transport layer is intended for the end-to-end transfer of data with between processes in end
systems. This basic multiplexing capability is addressed by TCP and UDP. If RTP were a transport
protocol it would unnecessarily duplicate this capability. By running RTP run on top of a transport
layer protocol such as UDP, the focus can be kept on the additional capabilities that are required to
provide real-time service.
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Communication Networks (2nd Edition)
Chapter 10 Solutions
10. 48. Can RTP be used to provide reliable real-time communication? If yes, explain how. If no, explain why not.
Solution:
RTP cannot be used to provide reliable real-time communication. It can only deal with reordering of
packets but cannot deal with packet losses. It can help to reduce the impact of losses and also delay
and jitter incurred in the network. RTCP can also be used to help the source to adjust to the
conditions of the network to reduce the impacts on loss, delay, and jitter. But nevertheless it cannot
guarantee a completely reliable real-time communication.
10.49. Can RTP operate over AAL5? Explain.
Solution:
RTP can operate over AAL5 by mapping RTP PDUs directly onto AAL5 PDU’s in point-to-point
applications. RTP is designed for multicast scenarios, which is not readily handled by ATM.
10.50. Suppose that the marker bit in the RTP header is used to indicate the beginning of a talk spurt. Explain the
role of the timestamps and sequence numbers at the receiver.
Solution:
The timestamps are used for synchronization and jitter cancellation. The sequence numbers are used
to detect packet losses, and to recover packet sequences. In the case of the talk spurt, the
timestamps and sequence numbers are used to regenerate the same talk spurt at the receiver with as
much accuracy as possible. The marker indicates the sequence of packets that contain the talkspurt.
10.51. Discuss the relative usefulness of RTCP for unicast and multicast connections.
Solution:
RTCP can be more useful in unicast connections because in this case the source should adjust to
only one route and one receiver. In the case of multicast connections RTCP will provide feedback on
multiple routes and if the source adjusts itself to any one of the routes other routes will share the
impact unnecessarily.
10.52. Run Microsoft NetMeeting or some other conferencing application that carries voice and/or video. Capture
the packets that are exchanged during the setting up of the session and during the data transfer phase. Determine
what protocols are used for session setup and for data transport. Examine details of the fields in each packet type.
What type of media encoding is used?
Solution:
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Communication Networks (2nd Edition)
Chapter 10 Solutions
Setup Capture:
TCP control packets
No. Time
Source
Destination
Protocol Info
6 18.904189 192.168.0.149
192.168.0.143
TCP 1748 > 1720 [SYN] Seq=1739645016 Ack=0 Win=16384 Len=0
7 18.905196 192.168.0.143
192.168.0.149
TCP 1720 > 1748 [SYN, ACK] Seq=4252100644 Ack=1739645017 Win=17520 Len=0
8 18.905366 192.168.0.149
192.168.0.143
TCP 1748 > 1720 [ACK] Seq=1739645017 Ack=4252100645 Win=17520 Len=0
H.225.0 setup
11 19.497846 192.168.0.149
192.168.0.143
H.225.0 CS: Setup-UUIE
12 19.769449 192.168.0.143
192.168.0.149
TCP [Desegmented TCP]
13 20.099818 192.168.0.149
192.168.0.143
TCP 1748 > 1720 [ACK] Seq=1739645217 Ack=4252100649 Win=17516 Len=0
Alerting
14 20.101044 192.168.0.143
192.168.0.149
H.225.0 CS: Alerting-UUIE
16 20.501086 192.168.0.149
192.168.0.143
TCP 1748 > 1720 [ACK] Seq=1739645217 Ack=4252100688 Win=17477 Len=0
20 29.091030 192.168.0.143
192.168.0.149
TCP [Desegmented TCP]
21 29.329256 192.168.0.149
192.168.0.143
TCP 1748 > 1720 [ACK] Seq=1739645217 Ack=4252100692 Win=17473 Len=0
Connecting
22 29.330385 192.168.0.143
192.168.0.149
H.225.0 CS: Connect-UUIE
23 29.400799 192.168.0.149
192.168.0.143
TCP 1749 > 1862 [SYN] Seq=1740980379 Ack=0 Win=16384 Len=0
24 29.401781 192.168.0.143
192.168.0.149
TCP 1862 > 1749 [SYN, ACK] Seq=4253464033 Ack=1740980380 Win=17520 Len=0
25 29.401944 192.168.0.149
192.168.0.143
TCP 1749 > 1862 [ACK] Seq=1740980380 Ack=4253464034 Win=17520 Len=0
26 29.405685 192.168.0.149
192.168.0.143
TCP [Desegmented TCP]
27 29.453530 192.168.0.143
192.168.0.149
TCP [Desegmented TCP]
28 29.453752 192.168.0.149
192.168.0.143
H.245 TerminalCapabilitySet MasterSlaveDetermination
29 29.455958 192.168.0.143
192.168.0.149
H.245 TerminalCapabilitySet MasterSlaveDetermination
Negotiating
30 29.465312 192.168.0.149
192.168.0.143
TCP [Desegmented TCP]
31 29.471165 192.168.0.143
192.168.0.149
TCP [Desegmented TCP]
channel usage
32 29.471402 192.168.0.149
192.168.0.143
H.245 TerminalCapabilitySetAck MasterSlaveDeterminationAck
33 29.472271 192.168.0.143
192.168.0.149
H.245 TerminalCapabilitySetAck MasterSlaveDeterminationAck
34 29.678682 192.168.0.149
192.168.0.143
TCP [Desegmented TCP]
35 29.679868 192.168.0.143
192.168.0.149
H.245 OpenLogicalChannel OpenLogicalChannel OpenLogicalChannel
Requesting
43 30.532924 192.168.0.149
192.168.0.143
TCP 1749 > 1862 [ACK] Seq=1740980892 Ack=4253464654 Win=16900 Len=0
bandwidth
45 35.545028 192.168.0.143
192.168.0.149
RTCP Receiver Report
48 35.546773 192.168.0.149
192.168.0.143
RTCP Receiver Report
50 1239.290373 192.168.0.149
192.168.0.143
RSVP PATH Message. SESSION: IPv4, Destination 192.168.0.143, Protocol 17, Port 49608.
SENDER TEMPLATE: IPv4, Sender 192.168.0.149, Port 49608.
51 1239.322672 192.168.0.143
192.168.0.149
RSVP RESV Message. SESSION: IPv4, Destination 192.168.0.143, Protocol 17, Port 49608.
Data Transfer Capture:
No. Time
Source
Destination
Protocol Info
62 1240.429251 192.168.0.143
192.168.0.149
TCP 1862 > 1749 [ACK] Seq=4253464654 Ack=1740980896 Win=17004 Len=0
RTP video
63 1240.429470 192.168.0.149
192.168.0.143
H.245 OpenLogicalChannelReject OpenLogicalChannelAck OpenLogicalChannelAck
H.263 packet
MiscellaneousCommand MiscellaneousCommand OpenLogicalChannelConfirm
65 1240.622962 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12977, Time=3017250
66 1240.623218 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12978, Time=3017250
71 1240.854456 192.168.0.143
192.168.0.149
TCP 1862 > 1749 [ACK] Seq=4253464654 Ack=1740980980 Win=16920 Len=0
72 1240.854730 192.168.0.143
192.168.0.149
TCP 1503 > 1752 [ACK] Seq=4255085782 Ack=1742545885 Win=17315 Len=0
74 1240.915746 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12980, Time=3030750
75 1240.916004 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12981, Time=3030750
76 1240.916239 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12982, Time=3030750, Mark
87 1240.977683 192.168.0.149
192.168.0.143
TCP 1753 > 1503 [PSH, ACK] Seq=1742695460 Ack=4255248829 Win=17520 Len=25
88 1240.979358 192.168.0.143
192.168.0.149
TCP 1503 > 1753 [PSH, ACK] Seq=4255248829 Ack=1742695485 Win=17495 Len=21
89 1241.212546 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12983, Time=3044070
More
90 1241.212801 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12984, Time=3044070, Mark
91 1241.231672 192.168.0.149
192.168.0.143
TCP 1753 > 1503 [ACK] Seq=1742695485 Ack=4255248850 Win=17499 Len=0
control
92 1241.231775 192.168.0.149
192.168.0.143
TCP 1749 > 1862 [ACK] Seq=1740980980 Ack=4253464658 Win=16896 Len=0
93 1241.232617 192.168.0.143
192.168.0.149
H.245 MiscellaneousCommand
97 1241.364228 192.168.0.143
192.168.0.149
RTP Payload type=ITU-T H.263, SSRC=4062428632, Seq=32093, Time=1885770, Mark
98 1241.366210 192.168.0.149
192.168.0.143
TCP [Desegmented TCP]
99 1241.380190 192.168.0.149
192.168.0.143
TCP 1753 > 1503 [PSH, ACK] Seq=1742695485 Ack=4255248850 Win=17499 Len=268
100 1241.384497 192.168.0.143
192.168.0.149
TCP 1503 > 1753 [PSH, ACK] Seq=4255248850 Ack=1742695753 Win=17227 Len=106
Time=51984, Mark
163 1242.823618 192.168.0.143
192.168.0.149
RTP Payload type=ITU-T G.723, SSRC=756814963, Seq=32091, Time=52704
RTP audio
164 1242.836785 192.168.0.149
192.168.0.143
TCP 1753 > 1503 [ACK] Seq=1742695794 Ack=4255249018 Win=17331 Len=0
G.723
165 1242.836872 192.168.0.149
192.168.0.143
TCP 1754 > 1503 [ACK] Seq=1742808026 Ack=4255332966 Win=17168 Len=0
166 1242.897739 192.168.0.143
192.168.0.149
RTP Payload type=ITU-T G.723, SSRC=756814963, Seq=32092, Time=53424
167 1242.908675 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12996, Time=3120210
168 1242.908963 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12997, Time=3120210
173 1243.099425 192.168.0.143
192.168.0.149
RTP Payload type=ITU-T H.263, SSRC=4062428632, Seq=32109, Time=1963890, Mark
174 1243.107713 192.168.0.143
192.168.0.149
RTP Payload type=ITU-T G.723, SSRC=756814963, Seq=32093, Time=54144
175 1243.202816 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=12999, Time=3133620
176 1243.203076 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=13000, Time=3133620
177 1243.205246 192.168.0.149
192.168.0.143
RTP Payload type=ITU-T H.263, SSRC=1488263488, Seq=13001, Time=3133620, Mark
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Communication Networks (2nd Edition)
Chapter 10 Solutions
10.53. Repeat the previous problem for a distributed network game.
Solution:
Pending
10.54. Suppose a router acts as a firewall and filters packets. Can the router identify RTP packets? Can the router
identify RTP packets from different connections?
Solution:
RTP uses randomly selected port number. Firewalls normally restrict all port numbers except for a
select set of well-known numbers. Consequently RTP packets can have difficulty traversing firewalls.
Various fixes for different types of RTP-based applications and firewalls are available.
10.55. Suppose you want to implement an IP telephony system.
Solutions follow questions:
(a) Look up the RTP payload format for G.711. Discuss the transmission efficiency for the resulting system.
G.711 is used for encoding the 64 Kbps voice. Assuming 10 msec sampling frame size, the payload
is 80 bytes RTP header is 40 bytes and the efficiency is 66%. Efficiency will increase if larger frame
size is used. For 20 msec frame size, payload is 160 bytes and efficiency is 80%. For 30 msec frame
size, payload is 240 bytes and efficiency is 85%.
(b) Repeat for G.729.
In this case, the signal rate is 8 kbps, and for 10 msec frame size the payload is only 10 bytes.
Considering 40 bytes RTP header, the efficiency is 20%. Assuming 20 msec frame size, payload is
20 bytes and the efficiency is 33%. Assuming 30 msec frame size, payload is 30 bytes and efficiency
is 43%.
10.56. Explain the relevance to scalability of the SIP protocol not having to maintain state. Contrast this with the
case of traditional telephone networks.
Solution:
In the telephone network once a call is setup the network is responsible to maintain the state of the
connection all across the network in order to keep the connection. In SIP protocol once the call is
setup the end systems can continue the call without requiring maintaining the connection. The two
end systems can communicate directly once they know each other’s address. As a result SIP does
not have scalability problem, any number of new users can join in and make any number of new calls.
The only limitation is the network traffic. In contrast, in the telephone network scaling the network
requires installing complex switches and other equipment proportional to the number of new users
and new calls they are required to handle.
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Communication Networks (2nd Edition)
Chapter 10 Solutions
10.57. Consider whether and how SIP can be used to provide the following services:
Solutions follow questions:
(a) Call Display: The number and/or name of the calling party is listed on a screen before the call is answered.
The invite message by the SIP client carries the caller information. The information can be extracted
by the user agent server and displayed on the screen when the message is received.
(b) Call Waiting: A special sound is heard when the called party is on the line and another user is trying to
reach the called party.
The user agent server in the called party’s system can indicate the new call request to the user by
generating a special sound whenever it receives the new invite message while another call is in
progress.
(c) Call Answer: If the called party is busy or after the phone rings a prescribed number of times, the network
gives the caller the option of leaving a voice message.
The called party’s proxy server or the user agent server of the called party can return a message to
the calling party giving the option of leaving a message in the case the called party is busy or does
not answer the call. An agent can be used to record the message. The agent can directly
communicate with the user agent client of the caller to record the message.
(d) Three-Way-Calling: Allows a user to talk with two other people at the same time,
SIP supports multi party conference calls. Three-way calling can be handled as a three-party
conference call. The user agent client of the caller can create two simultaneous invite messages to
the two other people to initiate the call. Once the call is setup, the messages of each party are sent to
both of the other two parties.
10.58. A professor has three locations at his university: his regular professorial office, his lab, and the
administrative office for a research center he operates.
Solutions follow questions:
(a) Explain how the fork capability of SIP may be used to direct an Internet phone call to the professor.
The addresses of the three locations are given to the proxy server managing the SIP calls for the
professor. The proxy server forks the invite messages that it receives to all three locations.
(b) Suppose that the professor spends Tuesdays working at a small startup, where he also has an Internet
phone. Explain how a REGISTER message can be used to direct calls from the professor’s office in the
university to the startup.
The register method is used to send location information to a SIP server. In this method the user
sends a REGISTER message to its SIP server giving it a temporary forwarding address. In this
example the professor initiates the method using the IP telephone. The IP telephone then sends the
register message to the SIP server automatically.
(c) Suppose that on Wednesday, the professor forgets to change the forwarding instructions from the day
before. Can a call placed to him at the startup reach him at the office? At the lab?
The calls placed at the startup cannot reach him at the office or lab until the proxy server locates him
in one of the new locations unless the call forwarding method described in (a) is in use.
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Communication Networks (2nd Edition)
Chapter 10 Solutions
10.59. Suppose that a university campus is covered by 802.11 wireless LANs that are connected to the campus IP
backbone. Explain how SIP, RTP and associated protocols can be used to provide mobile phone service across the
campus. Explain how the service can be extended to provide service to other universities.
Solution:
SIP and RTP operate in the same manner as they would in a wired network environment, and the
extension to other sites would follow the same approach as in wired networks. The tricky part is in
authentication, the handling of IP addressing, the interconnecting of the various wireless LANs, and
the handling of mobility and the handoff latency impacts on voice service. One extreme approach
would have each wireless access point connected to a router and all communications thereafter
handled in layer 3. Mobile IP could be used in this case to handle mobility. Current implementations
may introduce large handoff delays. An opposite extreme would have all wireless LANs
interconnected in a large layer-2 bridged network. Bridge learning algorithms could be used to
determine the location of a given user. The learning time of current implementations are too large for
effective handoff of voice calls, and so modifications of bridge learning algorithms may be required.
10.60. A recent proposal involves replacing the H.323 gateway with a trunking gateway that is responsible solely
for media format conversion. Signaling to establish a call is handled by a call agent that talks to the client on the
packet network side, to a signaling #7 gateway to the telephone network, and to the trunking gateway.
Solutions follow questions:
(a) Explain how a telephone call is set up between an Internet terminal and a telephone set.
Let’s assume that the client on the packet network initiates a call. The client is connected to the call
agent and sends messages indicating the request for the call and the desired destination. The call
agent talks to an SS7 gateway to perform the signaling to establish the call in the telephone network.
At the same time the call agent talks to the trunking gateway to setup a trunk for the call. Once the
signaling is done, the call is setup between the internet terminal and the telephone set through the
trunking gateway.
(b) Does this system have greater scalability than an H.323 gateway has?
H.323 involves complex signaling and requires heavy processing that can only be done by powerful
end systems attached to packet network. The new system described here is more scalable because it
eliminates the need for H.323 servers and replaces them with much simpler agents in the network.
(c) Can the call agents be used to move telephone intelligent network functionality to the Internet?
Yes the call agents can be used to provide some of the intelligent network functionality.
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